A crate of mass 5kg is pushed up a rough incline by means of a horizontal pushin

Question

A crate of mass 5kg is pushed up a rough incline by means of a horizontal pushing force of constant magnitude 50N. The incline makes an angle 30° with respect to the horizontal and has a coefficient of kinetic friction 0.2 with the surface of the crate. The crate moves a distance 3m along the incline. Find

a) the work done on the crate by gravity.
b) the work done on the crate by the pushing force.
c) the work done on the crate by the normal force.
d) the work done on the crate by friction.
e) the final speed of the crate if the crate started from rest.

[ Please use a starting equation with all steps included ]

1. A crate of mass 5kg is pushed up a rough incline by means of a horizontal pushing force of constant magnitude 50N. The incline makes an angle 30 with respect to the horizontal and has a coefficient of kinetic friction 0.2 with the surface of the crate. The crate moves a distance 3m along the incline. Find a) the work done on the crate by gravity. b) the work done on the crate by the pushing force. c) the work done on the crate by the normal force. d) the work done on the crate by friction e) the final speed of the crate if the crate started from rest. Fx (N)

Explanation / Answer

here,

mass of crate , m = 5 kg

constant force , F = 50 N

theta = 30 degree

uk = 0.2

s = 3 m

a)

the work done on the crate by gravity , Wg = m * g * sin(theta) * s * cos(180)

Wg = 5 * 9.81 * sin(30) * 3 * (-1)

Wg = - 73.5 J

b)

the work done on the crate by the pushing force , Wf = F * s

Wf = 50 * 3 J = 150 J

c)

the work done on the crate by the normal force , Wn = N * s * cos(90) = 0 J

d)

the work done on the crate by friction , Wff = uk * m * g * cos(theta) * s * cos(180)

Wff = 0.2 * 5 * 9.81 * cos(30)*3 * ( -1) J

Wff = - 25.4 J

e)

let the final speed be v

using work energy theorm

work done = kinetic energy gained

Wf + Wg + Wn + Wff = 0.5 * m * (v^2 - 0^2)

- 73.5 + 150 + 0 - 25.4 = 0.5 * 5 * v^2

solving for v

v = 4.52 m/s

the final speed is 4.52 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.