Question

(d) You observe someone pulling a block of mass 41 kg across a low-friction surface. While they pull a distance of 4 m in the direction of motion, the speed of the block changes from 5 m/s to 7 m/s. Calculate the magnitude of the force exerted by the person on the block.

Answer 1

Answer:

The magnitude of the force is 124.23 N.

Explanation:

To known the magnitude of the force is necessary to find the acceleration, that can be done by means of the equations for a a Uniformly Accelerated Motion:

$v_{f} = v_{i} + at$  (1)

Where $v_{f}$ is the final velocity, $v_{i}$ is the initial velocity, a is the acceleration and t is the time.  

$a = \frac{v_{f}-v_{i}}{t}$ (2)

$d = v_{i}t + \frac{1}{2}at^{2}$ (3)

By replacing (2) in equation (3) it is gotten:

$d = v_{i}t + \frac{1}{2}(\frac{v_{f}-v_{i}}{t})t^{2}$

$d = v_{i}t + \frac{1}{2}(v_{f}-v_{i})t$

$d = v_{i}t + \frac{1}{2}v_{f}t-\frac{1}{2}v_{i}t$

Therefore, by subtracting the first and third term ($v_{i}t-\frac{1}{2}v_{i}t$ ⇒ $\frac{1}{2}v_{i}t$) it is got:

$d = \frac{1}{2}v_{f}t + \frac{1}{2}v_{i}t$

Applying common factor for $\frac{1}{2}t$:

$d = \frac{1}{2}(v_{f} + v_{i})t$  (4)

Equation (4) can be rewritten in terms of t:

$t = \frac{2d}{(v_{f} + v_{i})}$

$t = \frac{(2)(4m)}{(7m/s + 5m/s)}$

$t = \frac{8m}{12m/s}$

$t = 0.66s$

By knowing the time is possible to determine the acceleration by means of equation (2):

$a = \frac{7m/s-5m/s}{0.66s}$  

$a = \frac{2m/s}{0.66s}$

$a = 3.03m/s^{2}$

The magnitude of the force exerted by the person on the block can be determine by means of Newton's second law:

$F = ma$ (5)

$F = (41Kg)(3.03m/s^{2})$

$F = 124.23 Kg.m/s^{2}$

$F = 124.23 N$

Hence, the magnitude of the force is 124.23 N.