Question

1.Calculate the work done by a boy pushing a 20 kg load for 10 m along a horizontal track with frictional force of a 20 N if the load accelarates at 20 m/s.

Answer 1

Given:

I - length of way,

v-final speed,

$\mathsf{F}_{\mathsf{fr}}$ -frictional force

m - mass of the load

From the second Newton law:

$$m \bar {a} = \sum \bar {F}$$

As we can see from the picture above, force (F) that pulls load has opposite direction to frictional force. So in projection on the axis, along which load moves:

$$m a = F - F _ {f r},$$

Or

$$F = F _ {f r} + m a$$

Where

$$a = \frac {v ^ {2}}{2 * l}$$

Job, done by boy:

$$A = \left(\bar {F} * \bar {l}\right)$$

Job is scalar product of the force vector and the displacement vector; in our case they are collinear.

Hence,

$$A = F * l = \left(F _ {f r} + m * \frac {v ^ {2}}{l}\right) * l = \left(2 0 + 2 0 * \frac {2 0 ^ {2}}{2 * 1 0}\right) * 1 0 = 4 2 0 0 J$$

(where J is Joules)