Work and Energy Situations Involving Kinetic Friction Problems and Solutions

Problem#1

A 40.0-kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between box and floor is 0.300, find (a) the work done by the applied force, (b) the increase in internal energy in the box-floor system due to friction, (c) the work done by the normal force, (d) the work done by the gravitational force, (e) the change in kinetic energy of the box, and (f) the final speed of the box.

Answer:

∑Fy = may = 0

n – mg = 0

n = mg = (40.0 kg)(9.80 m/s2) = 392 N, then

fk = µkn = 0.300(392 N) = 118 N

(a) the work done by the applied force is

WF = F∆r cosθ = (130 N)(5.00 m) cos00 = 650 J

(b) the increase in internal energy in the box-floor system due to friction,

∆Eint = fk∆x = (118 N)(5.00 m) = 588 J

(c) the work done by the normal force is

Wn = n∆x cos900 = 0

(d) the work done by the gravitational force is

Wg = mg∆x cos2700 = 0

(e) the change in kinetic energy of the box is

∆K = Kf – Ki = ∑W - ∆Eint

Kf – 0 = 650 J – 588 J + 0 + 0

Kf = 62.0 J

(f) the final speed of the box, given by

Kf = ½ mvf2

½ (40.0 kg)vf2 = 62.0 J

vf = 1.76 m/s

Problem#2

A 2.00-kg block is attached to a spring of force constant 500 N/m as in Figure 2. The block is pulled 5.00 cm to the right of equilibrium and released from rest. Find the speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface is 0.350.

Answer:

(a) the speed of the block as it passes through equilibrium if the horizontal surface is frictionless,

Ws = ½ kxi2 – ½ kxf2 = ½ mvf2 – ½ mvi2

½ (500 N/m)(5.00 x 10-2 m)2 – 0 = ½ (2.00 kg)vf2

0.625 J = vf2

vf = 0.791 m/s

(b) the speed of the block as it passes through equilibrium if the coefficient of friction between block and surface is 0.350,

∆K = Wc + Wnc

½ mvf2 – ½ mvi2  = Wc + (–fk∆x)

½ (2.00 kg)vf2 – 0 = 0.625 J + 0.350(2.00 kg)(9.80 m/s2)(5.00 x 10-2 m)

vf2 = 0.282 J

vf = 0.531 m/s

Problem#3

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 m. (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate–incline system due to friction. (c) How much work is done by the 100-N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5.00 m?

Answer:

(a) Work is done by the gravitational force on the crate is

Wg = mgdcos(90.00 + θ)

Wg = (10.0 kg)(9.80 m/s2)(5.00 m)cos1100 = –168 J

(b) The increase in internal energy of the crate–incline system due to friction is

∆Eint = fkd

with, fk = µkn = µkmgcosθ, then

∆Eint = µkmgd cosθ

∆Eint = (0.400)(10.0 kg)(9.80 m/s2)(5.00 m) cos20.00

∆Eint = 184 J

(c) Work is done by the 100-N force on the crate is

WF = Fd = (100 N)(5.00 m) = 500 J

(d) The change in kinetic energy of the crate is

∆K = ∑Wother - ∆Eint 

∆K = Wg + WF - ∆Eint 

∆K = –168 J + 500 J – 184 J = 148 J

(e) The speed of the crate after being pulled 5.00 m is

∆K = ½ mvf2 – ½ mvi2

148 J = ½ (10.0 kg)vf2 – ½ (10.0 kg)(1.50 m/s)2

159.25 J = (5.00 kg)vf2

vf = 5.64 m/s

Problem#4

A 15.0-kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0° above the horizontal. The block is displaced 5.00 m, and the coefficient of kinetic friction is 0.300. Find the work done on the block by(a) the 70-N force, (b) the normal force, and (c) the gravitational force. (d) What is the increase in internal energy of the block-surface system due to friction? (e) Find the total change in the block’s kinetic energy.

Answer:

From ∑Fy = may = 0, we use

n + Fsin20.00 – mg = 0

n = 70 kg(9.80 m/s2) – (70 N)sin20.00 = 123 N

(a) the work done on the block by the 70-N force is

W = F∆rcosθ = (70 N)(5.00 m)cos20.00 = 329 J

(b) the work done on the block by the normal force,

W = n∆rcosθ = (123 N)(5.00 m)cos90.00 = 0 J

(c) the work done on the block by the gravitational force is

W = mg∆rcosθ = (147 N)(5.00 m)cos(–90.00) = 0 J

(d) The increase in internal energy of the block-surface system due to friction is

∆Eint = fk∆x = µn∆x

∆Eint = (0.300)(123 N)(5.00 m) = 185 J

(e) the total change in the block’s kinetic energy given by

∆K = Kf – Ki = ∑W - ∆Eint

∆K = 329 N – 185 J = 144 J

Problem#5

A sled of mass m is given a kick on a frozen pond. The kick imparts to it an initial speed of 2.00 m/s. The coefficient of kinetic friction between sled and ice is 0.100. Use energy considerations to find the distance the sled moves before it stops.

Answer:

We use

∆K = ∑W = Wc + Wnc

Kf – Ki = Wc + Wnc

0 – ½ mvi2 = 0 + (–fk∆x)

½ mvi2 = µkmg∆x

½ (2.00 m/s)2 = (0.100)(9.80 m/s2)∆x

∆x = 2.04 m