Question

The figure below shows a mass m=1.0 kg which is initially moving with speed v=6.7 m/s at the top of a frictionless hill of height h=4.9 m. It slides down the hill until it encounters a flat section with unknown coefficient of kinetic friction µ_k. If the object travels a distance D=10.8 m before reaching a speed of v/2, what is the coefficient of kinetic friction µ_k?

✓ -D h

Answer 1

Given:

Mass of the object, $m=1kg$

Initial speed of the object,

V = 6.7m/s

Initial height of the object, $h=4.9m$

Distance traveled by the object on the rough flat surface, $D=10.8m$

Final speed,

vf 6.7m/s 2 3.35m/s

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Workdone by friction = Frictional force * Distance

Frictional force = Coefficient of friction*Normal force

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Consider the object

As the object slides down, its initial potential energy and initial kinetic energy is converted into final kinetic energy.

As the flat surface is rough, some portion of the initial energies will be lost due to friction.

$KE_{i}+PE_{i}=KE_{f}+W_{friction}$

$KE_{i}+PE_{i}=KE_{f}+F_{friction}*D$

$\frac{1}{2}mv^{2}+mgh=\frac{1}{2}mv_{f}^{2}+\mu _{k}N*D$

$\frac{1}{2}mv^{2}+mgh=\frac{1}{2}mv_{f}^{2}+\mu _{k}mg*D$

Cancel mass m

$\frac{1}{2}v^{2}+gh=\frac{1}{2}v_{f}^{2}+\mu _{k}g*D$

$0.5*(6.7m/s)^{2}+9.81m/s^{2}*4.9m=0.5*(3.35m/s)^{2}+\mu _{k}*9.81m/s^{2}*10.8m$

$0.5*6.7^{2}+9.81*4.9=0.5*3.35^{2}+\mu _{k}*9.81*10.8$

$22.445+48.069=5.61125+\mu _{k}*105.948$

$22.445+48.069-5.61125=\mu _{k}*105.948$

$64.90275=\mu _{k}*105.948$

$\frac{64.90275}{105.948}=\mu _{k}$

ANSWER: ${\color{Red} \mu _{k}=0.6126}$

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