A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.

Question

A crate of mass 10.0 kg is pulled up a rough incline with
an initial speed of 1.50 m/s. The pulling force is 100 N
parallel to the incline, which makes an angle of 20.0° with
the horizontal. The coefficient of kinetic friction is 0.400,
and the crate is pulled 5.00 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 100-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 5.00 m?

Explanation / Answer

Q is theta here, f is force of friction and F is pulling force and uk is coefficient of kinetic friction, vf is the final velocity and vi is the initial velocity a) Wg = m*g*h = 10*9.8*5*sin(20) = 167.6 J where h = d sin(Q) b) fiction work = uk*N*d but N = m*g*cos(Q) = 10*9.8*cos(20) = 92.1 N work = uk*N*d = .4*92.1*5 = 184.2 J c) W = F*d = 100*5 = 500 J this comes from the w = integral of F.dx even with the force of fiction above d) f = uk*N = uk*m*g*cos(Q) = 36.84 N F - m*g*sin(Q) - f = m*a solving for a we get a = 2.96 m/s^2 vf^2 = vi^2 + 2*a*d vf^2 = 31.85 change in kinetic energy = (1/2)(m)(vf^2-vi^2) = 148 J e) vf^2 = 31.85 from above Vf = 5.64 m/s

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