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 (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=(F)/(m)=(75 N )/(31.8 kg)\approx 2.36 (m)/(s^2)\\`

Now we can determine the displacement from the following formula:

`d = (1)/(2)a^2+v_0t+d_0`

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

`d = (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 = (v_f)/(a) = (15.2 (m)/(s))/(2.36 (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 = (1)/(2)at^2=(1)/(2)2.36 (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`