A force vec-F-3that-i-5hat-j- N acts on a particle whose position vector varies as vec-S-2t-2-hat-i-5hat-j- m- where t is time in seconds- The work done by this force from t-0 to t-2s is-23 Jzerocan-t be obtained32 J

Question

Toppr Admin · Accepted Answer

The work done by the-xA0-force -x2192-F-3t-i-5-j- is-xA0-W-x222B-x2192-F-d-x2192-s-x222B-3t-i-5-j-4tdt-i-xA0-x222B-2012t2dt-12-t3-203-12-8-x2212-0-3-32-xA0-J

A force $\to F = (3 t^i + 5^j)$ N acts on a particle whose position vector varies as $\to S = (2 t^{2}^i + 5^j)$ m, where $t$ is time in seconds. The work done by this force from $t = 0$ to $t = 2 s$ is:
23 J
zero
can't be obtained
32 J

The work done by the force $\to F = (3 t^i + 5^j)$ is

$W = \int \to F . d \to s = \int (3 t^i + 5^j) . (4 t d t^i)$

$= \int_{0}^{2} 12 t^{2} d t = \frac{12 [t^{3}]_{0}^{2}}{3} = \frac{12 (8 - 0)}{3} = 32 J$

A force →F=(3t^i+5^j) N acts on a particle whose position vector varies as →S=(2t2^i+5^j) m, where t is time in seconds. The work done by this force from t=0 to t=2s is:23 Jzerocan't be obtained32 J

The work done by the force →F=(3t^i+5^j) is W=∫→F.d→s=∫(3t^i+5^j).(4tdt^i) =∫2012t2dt=12[t3]203=12(8−0)3=32 J

A force $\to F = (3 t^i + 5^j)$ N acts on a particle whose position vector varies as $\to S = (2 t^{2}^i + 5^j)$ m, where $t$ is time in seconds. The work done by this force from $t = 0$ to $t = 2 s$ is:
23 J
zero
can't be obtained
32 J

The work done by the force $\to F = (3 t^i + 5^j)$ is

$W = \int \to F . d \to s = \int (3 t^i + 5^j) . (4 t d t^i)$

$= \int_{0}^{2} 12 t^{2} d t = \frac{12 [t^{3}]_{0}^{2}}{3} = \frac{12 (8 - 0)}{3} = 32 J$