Question

7. -10 points SerPSE10 8.3.P.007.MI. My Notes Ask Your Teacher A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.54 m/s. The pulling force is 104 N parallel to the incline, which makes an angle of 20.5° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.90 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 104-N force on the crate? CJ (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 4.90 m? m/s Need Help? Read It Master it

Answer 1

here,

the mass of crate , m = 9.2 kg

initial speed of crate , u = 1.54 m/s

force applied , F = 104 N

theta = 20.5 degree

coefficient of kinetic friction , uk = 0.4

s = 4.9 m

a)

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

Fg = - 9.2 * 9.81 * 4.9 * sin(20.5) J = - 154.9 J

b)

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

U = - 0.4 * 9.2 * 9.81 * 4.9 * cos(20.5) J

U = 165.7 J

c)

the work done by applied force ,Wf = F * s * cos(0)

Wf = 104 * 4.9 J= 509.6 J

d)

using Work-energy theorm

Work done by all forces = change in kinetic energy

Wf - U + Wg = change in kinetic energy

change in kinetic energy = 509.6 - 165.7 - 154.9 J

change in kinetic energy = 189 J

e)

let the final velocity be v

change in kinetic energy = 0.5 * m * ( v^2 - u^2)

189 = 0.5 * 9.2 * ( v^2 - 1.54^2)

solving for v

v = 6.6 m/s

the final velocity is 6.6 m/s