Question

A crate of mass 10.6 kg is pulled up a rough incline with an initial speed of 1.52 m/s. The pulling force is 94 N parallel to the incline, which makes an angle of 20.2° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.94 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 owing to friction.

(c) How much work is done by the 94-N force on the crate?

(d) What is the change in kinetic energy of the crate?

The increase in internal energy of the block corresponds to a
decrease in the kinetic energy. J

(e) What is the speed of the crate after being pulled 4.94 m?

Answer #1

here,

mass , m = 10.6 kg

initial speed , u = 1.52 m/s

pulling force , F = 94 N

theta = 22 degree

uk = 0.4

s = 4.94 m

a)

work done by the gravitational force on the crate , Wg = - m * g * s * sin(theta)

Wg = - 10.6 * 9.81 * 4.94 * sin(22) J

Wg = - 192.4 J

b)

the increase in internal energy , Ui = uk * m * g * cos(theta) * s

Ui = 0.4 * 10.6 * 9.81 * cos(22) * 4.94 J

Ui = 190.5 J

c)

the work done by 94 N force , Wf = F * s

Wf = 94 * 4.94 J = 464.36 J

d)

the change in kinetic energy of the crate, dKE = Wf + Wg - Ui

dKE = 464.36 - 192.4 - 190.5 J

dKE = 81.46 J

e)

let the final speed of crate be v

dKE = KEf - KEi

81.46 = 0.5 * 10.6 * ( v^2 - 1.52^2)

solving for v

v = 4.2 m/s

the final speed of the crate is 4.2 m/s

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