Practice # 8(1) (1)

PHY 101Practice Problems

Newton’s Laws of MotionSummer 2013

Instructor: Asad Hasan37) A 1 000-N crate is being pushed across a level floor at a constant speed by

a force F⃗ of 300 N at an angle of 20.0° below the horizontal, as shown in Figure P4.37a. (a) What is the coefficient of kinetic friction between the crate and the floor? (b) If the 300-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in Figure P4.37b, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in (a).

(a) Since the crate has constant velocity, .

Applying Newton’s second law:

, or

and , or

The coefficient of friction is then

k =0.256; a = 0.509 m/s2

(b) In this case,

so

The friction force now becomes

Therefore, and the acceleration is

41. The coefficient of static friction between the 3.00-kg crate and the 35.0°

incline of Figure P4.41 is 0.300. What minimum force F⃗ must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

Figure P4.41

The normal force acting on the crate is given by

. The net force tending to move the crate down

the incline is , where is the force of static friction between the crate and the incline. If the crate is in

equilibrium, then , so that

But, we also know

Therefore, we may write , or

F= 32.1 N

48. Objects of masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley as in Figure P4.48. The object m1 is held at rest on the floor, and m2 rests on a fixed incline of θ = 40.0°. The objects are released from rest, and m2 slides 1.00 m down the incline in 4.00 s. Determine (a) the acceleration of each object, (b) the tension in the string, and (c) the coefficient of kinetic friction between m2 and the incline.

Figure P4.48(a) Both objects start from rest and have accelerations of the same magnitude, a.

This magnitude can be determined by applying to the

motion of :

(b) Consider the free-body diagram of and apply Newton’s 2nd law:

or

a = 0.125 m/s2

T= 39.7 Nk = 0.235

(c) Considering the free-body diagram of :

or

so

Then

or

The coefficient of kinetic friction is

50. A 2.00-kg block is held in equilibrium on an incline of angle θ = 60.0° by

a horizontal force F⃗ applied in the direction shown in Figure P4.50. If the coefficient of static friction between block and incline is μs = 0.300, determine

(a) the minimum value of the force F⃗ to prevent the crate from sliding down the incline, and (b) the normal force exerted by the incline on the block.

When the minimum force is used, the block tends to

slide down the incline so the friction force, is directed up the incline.

While the block is in equilibrium, we have

(1)

and

(2)

F= 18.5 NN=25.8 N

For minimum F (impending motion), (3)

Equation (2) gives (4)

(a) Equation (3) becomes: , so Equation (1) gives

, or

(b) Finally, Equation (4) gives the normal force

53. In Figure P4.53, the coefficient of kinetic friction between the two blocks shown is 0.30. The surface of the table and the pulleys are frictionless. (a) Draw a free-body diagram for each block. (b) Determine the acceleration of the system. (c) Find the tension in the strings.

Figure P4.53

T1

T2

a = 5.7 m/s2

T1 =17 NT2 = 41 N

(a)

(b) For the 10-kg object:

or (1)

For the 2.0-kg object: or

so

Also, or (2)

Finally, for the 3.0-kg object:

or (3)

Substituting Equations (1) and (2) into Equation (3) yields:

or

(c) Substituting the computed value for the magnitude of the acceleration into

Equations (1) and (2) gives:

and

60. A 4.00-kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0° with the horizontal, as in Figure P4.60. The block accelerates to the right at 6.00 m/s2. Determine the coefficient of kinetic friction between block and ceiling.

Consider the vertical forces acting on the block:

,

so the normal force is

Now, consider the horizontal forces:

or

The coefficient of kinetic friction is then

63. A 2.00-kg aluminum block and a 6.00-kg copper block are connected by a light string over a frictionless pulley. The two blocks are allowed to move on a fixed steel block wedge (of angle θ = 30.0°) as shown in Figure P4.63. Making use of Table 4.2, determine (a) the acceleration of the two blocks and (b) the tension in the string.

Use: kAluminum -steel = 0.47; kCopper -steel = 0.36

Figure P4.63

k = 0.814

a = 0.232m/s2

T = 9.68 N

(a) Free-body diagrams for the two blocks are given at the right. The coefficient of kinetic friction for

aluminum on steel is while that for copper

on steel is . Since for each block,

and . Thus,

and

For the aluminum block:

giving (1)

For the copper block:

or (2)

Substituting Equation (1) into Equation (2) gives

or

(b) From Equation (1) above, 65. Two boxes of fruit on a frictionless horizontal surface are connected by a light string as in Figure P4.65, where m1 = 10 kg and m2 = 20 kg. A force of 50 N is applied to the 20-kg box. (a) Determine the acceleration of each box and the tension in the string. (b) Repeat the problem for the case where the coefficient of kinetic friction between each box and the surface is 0.10.

a) a = 1.7 m/s2

T=17 Nb) a = 0.69m/s2

T=17 N

Figure 1 is a free-body diagram for the system consisting of both blocks. The friction forces are

and . For this system, the tension in the connecting rope is an internal force and is not included in second law calculations. The second law gives

, which reduces to

(1)

Figure 2 gives a free-body diagram of alone. For this system, the tension is an external force and must be included in the second law. We find:

, or

(2)

(a) If the surface is frictionless, . Then, Equation (1) gives

and Equation (2) yields

(b) If , Equation (1) gives the acceleration as

while Equation (2) gives the tension as

69. Three blocks of masses 10.0 kg, 5.00 kg, and 3.00 kg are connected by light strings that pass over frictionless pulleys as shown in Figure P4.69. The acceleration of the 5.00-kg block is 2.00 m/s2 to the left, and the surfaces are rough. Find (a) the tension in each string and (b) the coefficient of kinetic friction between blocks and surfaces. (Assume the same μk for both blocks that are in contact with surfaces.)

Figure P4.69

The magnitude of the acceleration is for all three blocks and applying Newton’s second law to the 10.0-kg block gives

, or

Applying the second law to the 5.00-kg block gives:

With , this simplifies to: (1)

For the 3.00-kg block, the second law gives

With , this reduces to:

(2)

Solving Equations (1) and (2) simultaneously, and using the value of from above, we find that

,

,

(a) , , and (b)