Question

A particle of mass 0.01 kg travels along a space curve with velocity given by $4\hat{i}+16\hat{k}m/s$. After some time, its velocity becomes $8\hat{i}+20\hat{k}m/s$ due to the action of a conservative force. The work done on the particle during this interval of time is:

• A. $0.32J$
• B. $6.9J$
• C. $9.6J$
• D. $0.96J$

Answer 1

The correct option is D $0.96J$

Given that,

Initial velocity $v_1=4\hat{i}+16\hat{k}$

Final velocity $v_2=8\hat{i}+20\hat{k}$

Now, the kinetic energy is

Initial Kinetic Energy$K.E=\frac{1}{2}\times 0.01\times ({\left(4\right)}^2+{\left(16\right)}^2)$

Final Kinetic Energy $K.E=\frac{1}{2}\times 0.01\times ({\left(8\right)}^2+{\left(20\right)}^2)$

Now, work done is

Work done = final K.E – initial K.E

$W=\frac{1}{2}\times 0.01(64+400-16-256)$

$W=0.96J$

Hence, the work done is $0.96J$