Solved: "A 15.0 kg block is dragged over a..."

Khadija Wells

Answered question

2021-02-21

A 15.0 kg block is dragged over a rough, horizontal surface by a70.0 N force acting at 20.0 degree angle above the horizontal. The block is displaced 5.0 m, and the coefficient of kinetic friction is 0.3. Find the work done on the block by ; a) the 70.0 N force,b) the normal force, and c) the gravitational force. d) what is the increase in the internal energy of the block-surface system due to friction? e) find the total change in the kinetic energy of the block.

Answer & Explanation

delilnaT

Skilled2021-02-23Added 94 answers

1. the work done by the 70 N force is
$W_{1} = F d \cos θ = 70 \cdot 5 \cos 20 = 328.89 J$
2. The normal force produces no work because it is perpendicular to the motion's displacement.
$W_{2} = 0$
3. If the displacement is zero in the vertical plane, then the gravitational force is perpendicular to the displacement. so there was no work done.

Jeffrey Jordon

Expert2021-10-14Added 2605 answers

$W_{F} = F d = 70 \cdot 5 \cos 20 = 142.82 J$

$N = m g \cos x = 15 \cdot 9.8 \cos 90 = 0 J$

$W_{g} = - m g \cos 90 = 0 J$

$W_{f} = - u d (m g - F \sin x) = - 0.3 \cdot 5 \cdot (15 \cdot 9.8 - 71.9 \sin 20)$

$W_{f} = - 183.61 N$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Force, Motion and Energy

Didn't find what you were looking for?