Question

A block with a mass of 31.8 kg is pushed on a frictionless

Answer 1

Answer: Total work done on the block is 3670.5 Joules.

Step by step:

Work done:

$W = F.d.\cos\theta$

With F the force, d the displacement, and theta the angle of action (which is 0 since the block is pushed along the direction of displacement, and cos 0 = 1)

$W = F.d$

Given:

F = 75 N

m = 31.8 kg

Final velocity $v_f = 15.3 \frac{m}{s}$

In order to calculate the Work we need to determine the displacement, or distance the block travels. We can use the information about F and m to first figure out the acceleration:

$F = ma\\\implies a=\frac{F}{m}=\frac{75 N }{31.8 kg}\approx 2.36 \frac{m}{s^2}$

Now we can determine the displacement from the following formula:

$d = \frac{1}{2}a^2+v_0t+d_0$

Here, the initial displacement is 0 and initial velocity is also 0 (at rest):

$d = \frac{1}{2}at^2$

Now we still have "t" as unknown. But we are given one more bit of information from which this can be determined:

$v_f = a\cdot t_f\\\implies t_f = \frac{v_f}{a} = \frac{15.2 \frac{m}{s}}{2.36 \frac{m}{s^2}}\approx 6.44 s$

(using vf as final velocity, and tf as final time)

So it takes about 6.44 seconds for the block to move. This allows us to finally calculate the displacement:

$d = \frac{1}{2}at^2=\frac{1}{2}2.36 \frac{m}{s^2}\cdot 6.44^2 s^2 \approx 48.94 m$

and the corresponding work:

$W = F\cdot d=75 N\cdot 48.94 m =3670.5J$