A 15 kg block is initially moving along a smooth horizontal surface with a speed of v = 4 m/s to the left. It is acted by a force F, which varies in the manner shown. Determine the velocity of the block at t = 15 seconds. Given theat F=40 cos ((π/10))t

Question

A 15 kg block is initially moving along a smooth horizontal surface with a speed of v = 4 m/s to the left. It is acted by a force F, which varies in the manner shown. Determine the velocity of the block at t = 15 seconds. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ce38f8ea5d4ebc9d4c2fa9e989f94362-.jpeg" /> Given theat F=40 cos ((π/10))t (A) 12.5 m/s (B) 8.5 m/s (C) 20 m/s (D) 9.5 m/s

Tardigrade Admin · Accepted Answer

Change in linear momentum Δ p=∫ Fdt 15(vf+u)=∫ limits015 40cos ((π/10))t dt vf=-4+(40/25)[(sin (π/10)t/π/10)]015 =-4 4+(400/15 π)(-1)=-12.5 m s

Q. $A 15$ kg block is initially moving along a smooth horizontal surface with a speed of $v = 4$ m/s to the left. It is acted by a force $F$ , which varies in the manner shown. Determine the velocity of the block at $t = 15$ seconds.

Given theat $F = 40$ cos $(\frac{π}{10}) t$

$12.5$ m/s

$8.5$ m/s

$20$ m/s

$9.5$ m/s

Change in linear momentum $Δ p = \int F d t$
$15 (v_{f} + u) = 0 \int 15 40 cos (\frac{π}{10}) t d t$
$v_{f} = - 4 + \frac{40}{25} [\frac{s in ( π /10 ) t}{π /10}]_{_{0}}^{^{15}} = - 4$
$4 + \frac{400}{15 π} (- 1) = - 12.5 m s$

Q. A15 kg block is initially moving along a smooth horizontal surface with a speed of v=4 m/s to the left. It is acted by a force F, which varies in the manner shown. Determine the velocity of the block at t=15 seconds. Given theat F=40 cos (10π​)t

12.5 m/s

8.5 m/s

20 m/s

9.5 m/s