Question

The distance x moved by a body of mass 0.5 kg under the action of a force varies with time T as x(m)=3t²+4t+5 (here t is expresed in second). what is the work done by the force in first 2 second ?
a) 20 J
b) 40 J
c) 50 J
d) 60 J

Answer 1

Answer:

The work done by the force is 75 J.

Explanation:

Given that,

Mass = 0.5 kg

Time = 2 sec

Distance $x(m)=3t^2+4t+5$....(I)

We need to calculate the acceleration

The acceleration is the second derivative of the position of the particle.

$a=\dfrac{d^2x}{dt^2}=6\ m/s^2$

We need to calculate the distance traveled in 2 sec

Put the value of t in equation (I)

$x=3\times2^2+4\times2+5$

$x=25\ m$

We need to calculate the work done

Using formula of work done

$W=F\times x$

$W=ma\times x$

Where, m = mass

a = acceleration

x = distance

Put the value into the formula

$W=0.5\times6\times25$

$W=75\ J$

Hence, The work done by the force is 75 J.

Answer 2

Answer: 60J

Explanation: