Question

Constants A crate of fruit with a mass of 39.0 kg and a specific heat capacity of 3550 J/(kg · K) slides 7.70 m down a ramp inclined at an angle of 37.0 degrees below the horizontal. Part A If the crate was at rest at the top of the incline and has a speed of 2.50 m/s at the bottom, how much work Wf was done on the crate by friction? Use 9.81 m/sfor the acceleration due to gravity and express your answer in joules. ► View Available Hint(s) 150 AC * O ? Wf = Submit

Answer 1

here,

the mass of crate , m = 39 kg

the distance traveled , s = 7.7 m

theta = 37 degree

the speed at bottom , v = 2.5 m/s

let the work done by friction be Wf

using Work energy theorm

Wf + Wg = 0.5 * m * ( v^2 - u^2)

Wf + m * g * s * sin(theta) = 0.5 * 39 * ( 2.5^2 - 0^2)

Wf + 39 * 9.81 * 7.7 * sin(37) = 0.5 * 39 * 2.5^2

Wf = - 1651 J

the work done by friction on crate is - 1651 J