A box of mass 2 kg is released from rest at the top of the ramp shown in the fig

Question

A box of mass 2 kg is released from rest at the top of the ramp shown in the figure. The friction coefficient of the ramp is 0.1, the friction coefficient of the level surface is 0.05, and we use g = 10 m/s2.

The numbered circles indicate relevant situations for the problem: (1) when the box is at the top of the ramp, (2) when the box is at the bottom of the ramp (we assume a smooth transition from the ramp to the horizontal floor), and (3) at a distance s from the end of the ramp.
(a) Which of the following principle(s) apply when the box slides down the ramp?
(i) Wnet = K2 - K1 (ii) U1 + K1 = U2 + K2 (iii) U1 + K1 = U2 + K2 - Wfr
[1] only (i) [2] only (ii) [3] only (iii) [4] (i) and (ii) [5] (i) and (iii)
(b) Calculate the work done by the friction force as the box slides down the ramp. The work due to friction is
[1] -23 J [2] -20 J [3] -12 J [4] 20 J [5] 120 J
(c) The speed of the box in situation (2), at the bottom of the ramp, is 10 m/s. What is the kinetic energy of the box in state (2)?
[1] 10 m/s [2] 10 J [3] 100 m2/s2 [4] 100 J [5] 120 J
(d) Calculate the speed of the box in situation (3), after the box has traveled a distance of 19 m on a level surface with friction coefficient 0.05
[1] 0 m/s [2] 2 m/s [3] 4.5 m/s [4] 9 m/s [5] 10 m/s

A box of mass 2 kg is released from rest at the top of the ramp shown in the figure. The friction coefficient of the ramp is 0.1, the friction coefficient of the level surface is 0.05, and we use g = 10 m/s2.

Explanation / Answer

Here,

m = 2 Kg

coefficient of ramp , u1 = 0.1

coefficient of friction level surface ,u2 = 0.05

g = 10 m/s^2

a) Using conservation of energy and work energy theorum

the correct statement is

Wnet = K2 - K1

U1 + K1 = U2 + K2 - Wfr

the correct option is 5) i and iii

b)

theta = arctan(6/10) = 31 degree

work done by friction force on ramp = - u1 * m * g * cos(theta) * d

work done by friction force on ramp = - 0.1 * 2 * 9.8 * cos(31) * sqrt(6^2 + 10^2)

work done by friction force on ramp = - 19.6 J

c)

let the Kinetic energy at 2 is

KE2 = 0.5 * m * v^2

KE2 = 0.5 * 2 * 10^2

KE2 = 100 J

the energy of box in state 2 is 100 J

d)

let the speed of box at 3 is v3

Using work energy theorum

- 100 + 0.5 * 2 * v3^2 = - 0.05 * 2 * 9.8 * 19

sovling for v3

v3 = 9 m/s

the speed of box at 3 is 9 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.