A crate of mass 9.0 kg is pulled up a rough incline with an initial speed of 1.4

Question

A crate of mass 9.0 kg is pulled up a rough incline with an initial speed of 1.44 m/s. The pulling force is 110 N parallel to the incline, which makes an angle of 19.40 with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.02 m. (a) How much work is done by the gravitational force on the crate? ) Determine the increase in internal energy of the crate-incline system owing to friction. (c) How much work is done by the 110-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 5.02 m? m/s

Explanation / Answer

a) Workdone by the gravitationsl force = -m*g*h

= -m*g*d*sin(theta)

= -9*9.8*5.02*sin(19.4)

= -147 J

b) Increase is internal energy = -Workdone by friction

= -(kf*d*cos(180))

= fk*d

= mue_k*N*d

= mue_k*m*g*cos(theta)*d

= 0.4*9*9.8*cos(19.4)*5.02

= 167 J

c) Workdone by 110N = F*d

= 110*5.02

= 552 J

d) change in kinetic energy = Net workdone

= W_applied force + Wfriction + W_gravity

= 552 - 167 - 147

= 238 J

e) now use Work-energy theorem,

Wnet = (1/2)*m*(vf^2 - vi^2)

vf^2 = (2*Wnet/m + vi^2)

= sqrt(2*Wnet/m + vi^2)

= sqrt(2*238/9 + 1.44^2)

= 7.41 m/s

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