Question #283443

A 12N object is attached vertically by a string to a frictionless car that runs along a level track. The car with mass (m) kg covers a distance of 1.25m in 5s after it starts from rest. a)Find the final speed of the car. b)Find the loss in potential energy of 12 N object in falling through distance of 1.25m c) Gain in kinetic energy of the car and the hanging object. d)Mass of the car

Expert's answer

a) Find the final speed of the car.


v=2dt=0.5 m/s.v=\dfrac{2d}t=0.5\text{ m/s}.

b) Find the loss in potential energy of 12 N object in falling through distance of 1.25 m.


ΔE=mgh=121.25=15 J.\Delta E=mgh=12·1.25=15\text{ J}.

c) Gain in kinetic energy of the car and the hanging object.


ΔKE=12mcv2+12Wg2gh=(0.125mc+15) J.\Delta KE=\dfrac12m_cv^2+\dfrac12 \dfrac Wg·2gh=(0.125m_c+15)\text{ J}.


d) Mass of the car:


mc=m2d/t2g2d/t2=12.6 gm_c=m\dfrac{2d/t^2}{g-2d/t^2}=12.6\text{ g}


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