Question

a box of mass 6kg is accelerated from rest by a force across a floor at rate of 8ms-2 for 10s .find the net work done on the box
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Answer 1

Answer:

19200J

Explanation:

for calculation of work ,we need values of force and distance covered.

force is calculated by multiplying mass with acceleration.

distance is calculated by using equation of linear motion as shown in image

Answer 2

$\dag \: \underline{\sf AnsWer :}$

Here we are given a box whose mass (m) is 6 kg, the acceleration (a) is 8 m/s² and time interval is 10 second. We are asked to find the net work done (W). First of all we need to find the net force. To calculate net force we will use the formula of Newton's second law which is given below :

• Force = mass × acceleration

Now,simply plug in the given values in above formula :

$:\implies\sf Force = mass \times acceleration$

$:\implies\sf Force = 6 \times 8$

$:\implies \underline{ \boxed{\sf Force = 48 \: N}}$

• We have find the net force now we need to calculate the displacement (s) of the box by using second kinematical equation of motion :

$\dashrightarrow\:\:\sf s = ut + \dfrac{1}{2} at^2$

$\dashrightarrow\:\:\sf s = 0 \times 40 + \dfrac{1}{2} \times 8 \times (10)^2$

$\dashrightarrow\:\:\sf s =\dfrac{1}{2} \times 8 \times 100$

$\dashrightarrow\:\:\sf s = 4 \times 100$

$\dashrightarrow\:\: \underline{ \boxed{\sf s = 400 \: m}}$

• Now, we have calculated the force and displacement so we can find the net work done of the box. So, Let's do it :

$\longrightarrow\:\:\sf Work \: done = Fs \cos(\theta)$

$\longrightarrow\:\:\sf Work \: done = 48 \times 400 \cos(0 ^{ \circ})$

$\longrightarrow\:\:\sf Work \: done = 48 \times 400 \times 1$

$\longrightarrow\:\: \underline{ \boxed{\sf Work \: done = 19200 \: J}}$