Question

A crate of mass 10.3 kg is pulled up a rough incline with an initial speed of 1.43 m/s. The pulling force is 93.0 N parallel to the incline, which makes an angle of 19.4° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.97 m.

(a) How much work is done by the gravitational force on the crate?
J
(b) Determine the increase in internal energy (related to thermal energy, having the opposite sign of the work done by friction) of the crate-incline system due friction .
J
(c) How much work is done by the 93.0 N force on the crate?
J
(d) What is the change in kinetic energy of the crate?
J
(e) What is the speed of the crate after being pulled 4.97 m?
m/s

Answer 1

(a) Work done by the gravitational force on the crate W = gravitational force along the incline* displacement = mgsinθ*(-S) = 10.3*9.8*sin19.4*-4.97 = -116.63J

(b) Increase in internal energy = -ve of work done by friction

work done by friction = Frictional force* (-displacement) = normal reaction force * coefficient of friction * (-S) = mgcos19.4*μ*-4.97 = -189.47 J

Increase in internal energy= 189.47 J

(c) Work done by the 93.0 N force on the crate = F*S
= 93*4.97
= 462.2 J
So, The work done by the Force is 462.2 J

(d) Inital K.E.= m*Vi^2/2 = 10.3*1.43^2 / 2 = 10.53J

Now, net upward force along the incline on the crate is Rf = applied force- gravitational force component downwards the incline - frictional force dowawards the incline = F - mgsinθ - μmgcosθ = 93-10.3*9.8*sin19.4- 0.4*10.3*9.8*cos19.4 = 21.3N

Acceleration a due to this force = Rf/ m= 2.07 m/s^2

so final velocity v is found by Vf= (Vi^2+2*a*S)^1/2 = 4.7m/s

So, final K.E. = mVf^2/2 = 116.5 J

So, change in kinetic energy of the crate = final - initial K.E. = 116.5 - 10.53 = 106 J

(e) speed of the crate after being pulled 4.97 m = Vf = 4.7 m/s.

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