A girl of mass 35 , kg sits on a trolley of m | vedclass

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Question

(A) To find the work done on the trolley,we use the Work-Energy Theorem: $W = \Delta KE = KE_f - KE_i$. Given: Total mass $m = 35 \, kg + 5 \, kg = 40

Vedclass · Accepted Answer

Work done on the trolley is $-320 \, J$

Vedclass · Answer

(A) To find the work done on the trolley,we use the Work-Energy Theorem: $W = \Delta KE = KE_f - KE_i$. Given: Total mass $m = 35 \, kg + 5 \, kg = 40 \, kg$,initial velocity $u = 4 \, m/s$,final velocity $v = 0 \, m/s$. $W = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 = 0 - \frac{1}{2} 	imes 40 	imes (4)^2 = -20 	imes 16 = -320 \, J$. The work done on the trolley is $-320 \, J$. $(b)$ Since the girl is sitting on the trolley and not applying any additional force to change the motion relative to the trolley,the work done by the girl is $0 \, J$.

$A$ girl of mass $35 \, kg$ sits on a trolley of mass $5 \, kg$. The trolley is given an initial velocity of $4 \, m \, s^{-1}$ by applying a force. The trolley comes to rest after covering a distance of $16 \, m$. $(a)$ What is the work done on the trolley? $(b)$ What is the work done by the girl?

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