Question

A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140 N. Fp is parallel to the displacement of the block. The final speed of the block is 2.35 m/s.

a) How much work was converted to thermal energy? What work did friction do on the box?
b) What is the coefficient of friction?

Answer 1

The answer is "151.25 J and -547.64 J".

Step-by-step explanation:

`u = 0\\\\v = 2.35\ (m)/(sec)\\\\d = 5.0 \ m\\\\`

Using formula:

`v^2 = u^2 + 2 * a * d\\\\2.35^2 = 0^2 + 2 * a * 5\\\\a = (2.35^2)/(10) \\\\`

`= 0.55 \ (m)/(sec^2)\\\\`

`F_(net) = m * a\\\\F_(net) = 55 * 0.55 = 30.25\ N\\\\`

Calculating the Work by net force

`W = F_(net)* d\\\\W = 30.25 * 5 = 151.25 \ J\\\\`

The above work is converted into thermal energy.

Now,

`F_(net) = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k* m * g = u_k * 55 * 9.81\\\\F_f = 539.55 * u_k\\\\30.25 = 140 - u_k * 55 * 9.81\\\\u_k = ((140 - 30.25))/((55* 9.81))\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f * d\\\\W_f = -0.203 * 55 * 9.81 * 5\\\\Work\ done\ by\ friction = -547.64 \ J`