The kinetic energy of a 500 gram stone is 100 J. Against a force of 50 N, how long will it travel ?

Question

The kinetic energy of a 500 gram stone is 100 J. Against a force of 50 N, how long will it travel ? (A) 0.2 s (B) 0.1 s (C) 0.3 s (D) 0.4 s

Tardigrade Admin · Accepted Answer

Here, m = 500 g = 0.5 kg KE = 100 J, F = 50 N a=-(F/m)=-(50/0.5)=-100 m s-2 As (1/2)mu2=KE ∴ u=√(2 kE/m)=√(2×100/0.5)=20 m s-1 and upsilon = 0 As upsilon=u+at 0=20-100t (∵ upsilon =0) t=0.2 s

Q. The kinetic energy of a 500 gram stone is 100 J. Against a force of 50 N, how long will it travel ?

$0.2 s$

$0.1 s$

$0.3 s$

$0.4 s$

Here, $m = 500 g = 0.5 k g$
$K E = 100 J, F = 50 N$
$a = - \frac{F}{m} = - \frac{50}{0.5} = - 100 m s^{- 2}$
As $\frac{1}{2} m u^{2} = K E$
$∴ u = \frac{2 k E}{m} = \frac{2 \times 100}{0.5} = 20 m s^{- 1}$
and $υ = 0$
As $υ = u + a t$
$0 = 20 - 100 t (∵ υ = 0)$
$t = 0.2 s$

Q. The kinetic energy of a 500 gram stone is 100 J. Against a force of 50 N, how long will it travel ?

0.2s

0.1s

0.3s

0.4s

Here, m=500g=0.5kg KE=100J,F=50N a=−mF​=−0.550​=−100ms−2 As 21​mu2=KE ∴u=m2kE​​=0.52×100​​=20ms−1 and υ=0 As υ=u+at 0=20−100t(∵υ=0) t=0.2s

$0.2 s$

$0.1 s$

$0.3 s$

$0.4 s$

Here, $m = 500 g = 0.5 k g$
$K E = 100 J, F = 50 N$
$a = - \frac{F}{m} = - \frac{50}{0.5} = - 100 m s^{- 2}$
As $\frac{1}{2} m u^{2} = K E$
$∴ u = \frac{2 k E}{m} = \frac{2 \times 100}{0.5} = 20 m s^{- 1}$
and $υ = 0$
As $υ = u + a t$
$0 = 20 - 100 t (∵ υ = 0)$
$t = 0.2 s$