05 February 2013

Physics Homework Ch4

Don't forget to +1, Like the videos, and share on Facebook

1. A football punter accelerates a football from rest to a speed of 13 m/s during the time in which his toe is in contact with the ball (about 0.24 s). If the football has a mass of 0.50 kg, what average force does the punter exert on the ball?
 __________ N

Try out the calculator (Must be using Chrome) or download the Calculator for this problem

2. A 11.0-kg object undergoes an acceleration of 2.9 m/s2.
a). What is the magnitude of the resultant force acting on it?
_________ N
b). If the same force is applied to a 22.0-kg object, what acceleration is produced?
________ m/s2.

3. One or more external forces are exerted on each object enclosed in a dashed box shown in the figure below. Identify the reaction to each of these forces. (Assume that (a), (b), and (c) occure on Earth while (d), (e), and (f) occur far from any outside gravitational influences. Select all that apply).

A)  The spring exerts a force B) The wagon exerts a force
downward on the Earth downward on the Earth due to contact
to the right on the hand upward on the Earth due to contact
upward on the Earth downward on the Earth due to gravitational attraction
to the right on the wall downward to the left on the handle
to the left on the hand upward on the Earth due to gravitational attraction
to the left on the wall upward to the left on the handle
C) the ball exerts a force  D) m exerts a force
downward to the left on the player to the left on M
upward to the right on the player to the right on M
downward on the Earth upward on M
downward to the right on the player downward on M
upward on the Earth
E) -q exerts a force F) The iron exerts a force
to the right on Q downward on the magnet
upward on Q upward on the magnet
to the left on Q to the left on the magnet
downward on Q to the right on the magnet

4. A bag of sugar weighs 6.00 lb on Earth. What would it weigh in newtons on the moon, where the free-fall acceleration is one-sixth that on Earth?

Repeat for Jupiter, where g is 2.64 times that on Earth.

Find the mass of the bag of sugar in kilograms at each of the three locations
Earth ____ kg
Moon ____kg

5. A freight train has a mass of 1.3 X 107 kg. If the locomotive can exert a constant pull of 7.4 X 105 N, how long does it take to increase the speed of the train from rest to 78 km/h?

6. A 73-kg man standing on a scale in an elevator notes that as the elevator rises, the scale reads 842 N. What is the acceleration of the elevator?
_____m/s2 upward.

7. A 5.6-g bullet leaves the muzle of a rifle with a speed of 320 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.75-m-long barrel of the rifle?

8. A boat moves through the water with two forces acting on it. One is a 1,500-N forward push by the water on the propeller, and the other is an 1,200-N resistive force due to the water around the bow.
a). What is the acceleration of the 800-kg boat?

b). If it starts from rest, how far will the boat move in 5.0 s?

c). What will its velocity be at the end of that time?

9. A 891-kg car starts from rest on a horizontal roadway and accelerates eastward for 5.00 s when it reaches a speed of 30.0 m/s. What is the average force exerted on the car during this time?
_______N eastward

10. A 2.00 kg block is held in equilibrium on an incline of angle θ = 65° by a horizontal force vector F  applied in the direction shown in the figure below. If the coefficient of static friction between block and incline is μs = 0.300, determine the following.

a). The minimum value of vector F
________ N
b). the normal force exerted by the incline on the block

Purchase the Excel Calculator For Problem #10 ($1.35), to solve quickly and easily

NOTE: I recommend you watch the video if you're having problems, but the calculator that I built will be able to get you at the right solution if you're in a hurry.
When you pay on paypal, click "back to mike b" and it will automatically download the file.

11. Consider the figure below.

a). Find the tension in each cable supporting the 700-N cat burglar. (Assume the angle θ of the inclined cable is 32.0° )
Inclined cable        ________N
Horizontal Cable   ________N
Vertical Cable       ________N

b). Suppose the horizontal cable were reattached higher up on the wall. Would the tension in the inclined cable
__or stay the same?

12. The leg and cast in the figure below weigh 347 N(w1). Determine the weight w2 and angle α needed so that no force is exerted on the hip joint by the leg plus cast.
w2 = _______ N
α =  _________ °

13. A crate of mass m = 25 kg rides on the bed of a truck attached by a cord to the back of the cab as in the figure. The cord can withstand a maximum tension of 63 N before breaking. Neglecting friction between the crate and truck bed, find the maximum acceleration the truck can have before the cord breaks.

14. Two packing crates of masses m1 = 10.0 kg and m2 = 4.00 kg are connected by a light string that passes over a frictionless pulley as in the figure. The 4.00-kg crate lies on a smooth incline of angle 34.0°

a). Find the acceleration of the 4.00-kg crate.
________m/s2  (up the incline)

b). Find the tension in the string
_________ N

15. Assume the three blocks (m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 4.0 kg) portrayed in the figure below move on a frictionless surface and a force F = 44 N acts as shown on the 4.0-kg block.

a) determine the acceleration given this system
 ________m/s2 (to the right)

b). Determine the tension ion the cord connecting the 4.0-kg and the 1.0-kg blocks

c). Determine the force exerted by the 1.0-kg block on the 2.0-kg block.

16. A block of mass m = 5.3 kg is pulled up a θ = 26° incline as in the figure with a force of magnitude F = 33 N.

a). Find the acceleration of the block if the incline is frictionless

b). Find the acceleration of the block if the coefficient of kinetic friction between the block and the incline is 0.12

17. A dockworker loading crates on a ship finds that a 17-kg crate, initially at rest on a horizontal surface, requires a 73-N horizontal force to set in motion. However, after the crate is in motion, a horizontal force of 52 N is required to keep it moving with a constant speed. Find the coefficient of static and kinetic friction between the crate and floor.
μstatic _______μkinetic ______

18. In the figure below, m1 = 9.5 kg and m2 = 3.5 kg. The coefficient of static friction between m1 and the horizontal surface is 0.5, and the coefficient of kinetic friction is 0.30.

a) If the system is released from rest, what will its acceleration be?
_________m/s2b) If the system is set in motion with m2 moving downward, what will be the acceleration of the system?

19. A 1,400-N crate is being pushed across a level floor at a constant speed by a force vector F  of 330 N at an angle of 20.0° below the horizontal, as shown in the figure (a) below.

a). What is the coefficient of kinetic friction between the crate and the floor?
μk = ________

b). If the 330-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in the figure (b), what will be the acceleration of the crate?
Assume that the coefficient of friction is the same as that found in part (a).

20. The coefficient of static friction between the m = 3.30-kg crate and the 35.0° incline of the figure below is 0.295. What minimum force vector F  must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?



  1. I love the song break after number 13!!!

  2. I was wondering who would stick around and listen to it! Did you see any special names appear on screen?

  3. I thought gravity was -9.8?

  4. Gravity is -9.8m/s^2;
    Now, saying that, here is the caveat.
    It can be considered a +9.8 as well, depending on how you establish your vectors.
    An example would be to consider a hole in the ground. If we say that moving into the hole is increasing our distance from earth's surface, that would imply that moving upwards is going at a negative distance from the surface. In both of these cases we would consider gravity to be +9.8; because the direction of force is the direction of positive displacement.

  5. Also, I'm not sure what question you're asking about, but I would also add this:
    in the eqyation Delta X = Vi(t)+(1/2)a(t^2),
    If a = gravity, then you may see me skip a step on the arithmetic and simply put 1/2a as -4.9

    LMK If this helps.

  6. On question #12 you got a different answer because at about 12:00 your 84.265 morphed into 84.765 due to mad hand writing.. I should probably get an award for noticing.

    1. Yeah, and at 14:35 the answer comes up and I'm like, "that's not the answer I got the first time I worked this out... But I think it was a rounding error"
      NOPE! Just a handwriting error!

      2 Life Points for Sara

      NOTE: You can spend life points only at MOB University and they are traded for free access to physics videos.

  7. how did you get 5.4 on question 12 i can not figure it out

  8. Replies
    1. Kenneth...
      I typed the complete answer below. Hope this helps

  9. Kenneth Borger,
    I. The simple answer is: Use newtons second law to set up two equations:
    1. Equation for m1
    (m1)g - T = (m1)a
    2. Equation for m2
    T - (m2)sin(theta) = (m2)a

    II. Then you add the two equations together to cancel out the "tension"

    (m1)g - (m2)g * sin(theta) = (m1 + m2)a

    III. Solve for acceleration and then plug in your numbers.

    ((m1)g - (m2)g * sin(theta)) / (m1 + m2) = a

    ((10 X 9.8) - (4 X 9.8 X sin(34))) / (10 + 4) = 5.43426

    If you're getting an answer of 5.518568 then you have your calculator set to RADIANS instead of DEGREES.

    (34 radians = 1948.057 degrees). Since 34 Radians is like going around a complete circle 5.411268 times, then you'd actually be calculating 148 degrees.... And because the math for 148 degrees is the same as 32 degrees, you get pretty close to the same answer, in this case (Usually doesn't work out like that).

  10. On number 18, why do we subtract the normal force of m2 with the kinetic friction of m1's normal force? Because it is the object being subjected to the friction, I know, but why are we subtracting the kinetic friction from the normal force of m2?

    1. Jordan, I'm not sure I follow you. m2 does not have a normal force. The normal force is equal to the force that is being applied by the ground. m2 is suspended in the air, therefore there is no normal force being applied.

      You may be asking why we subtracted the force exerted by m1 from the force exerted by m2.
      - in this case, m1 is providing only a force via friction
      - m2 is exerting a force via mass and gravity

      Why I'm subtracting is, I'm following newtons law
      - The sum of all forces is equal to the total mass multiplied by the acceleration
      -- (Sum of all forces) = (m1+m2)a

      Since the forces are in opposing directions (vectors), it's customary to define one of the forces as a negative force and the other as a positive force. I was not clear on this step, so I apologize

      I'm simply defining the direction of movement as the positive vector, so the force applied by friction (in the opposite direction) would be the negative vector