Answer to Question #166561 in Physics for David Samson

Question #166561

a block of 1.5 kg is pushed up a rough 25 degrees inclined plane by a constant force of 15 newton acting parallel to the inclined plane. if the coefficient of the kinetic friction is 0.3 and the block is displaced by a distance 10cm along the inclined plane, calculate the work done against the force of gravity and against the force of friction


1
Expert's answer
2021-02-25T11:26:21-0500

a) Work done against the force of gravity can be found as follows:


Wg=mgh.W_g=mgh.

We can find hh from the geometry:


sinθ=hd,sin\theta=\dfrac{h}{d},h=dsinθ.h=dsin\theta.

Finally, we get:


Wg=mgdsinθ=1.5 kg9.8 ms20.1 msin25=0.62 J.W_g=mgdsin\theta=1.5\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot0.1\ m\cdot sin25^{\circ}=0.62\ J.

b) We can find the work done against the force of friction as follows:


Wfr=Ffrd=mgdμcosθ,W_{fr}=F_{fr}d=mgd \mu cos\theta,Wfr=1.5 kg9.8 ms20.1 m0.3cos180=0.441 J.W_{fr}=1.5\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot0.1\ m\cdot0.3\cdot cos180^{\circ}=-0.441\ J.

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