Question

A block of mass 20.0 kg slides from rest down a slope 2.00 meters long which is inclined at 37.6° to the horizontal. The coefficient of kinetic friction is 0.343. What is the speed of the block at the bottom of the slope?
Select one:
a. 4.33 m/sec
b. 5.57 m/sec
c. 2.86 m/sec
d. 3.14 m/sec
e. 3.64 m/sec

Answer 1

Answer on Question #63310, Physics / Mechanics | Relativity

A block of mass $20.0\mathrm{kg}$ slides from rest down a slope 2.00 meters long which is inclined at $37.6{}^{\circ}$ to the horizontal. The coefficient of kinetic friction is 0.343. What is the speed of the block at the bottom of the slope?

Select one:

a. $4.33\mathrm{m / sec}$

b. $5.57\mathrm{m / sec}$

c. $2.86\mathrm{m / sec}$

d. $3.14\mathrm{m / sec}$

e. $3.64\mathrm{m / sec}$

Solution:

$F_{x}$ and $F_{y}$ are components of weight, $F_{g}$ ; $F_{N}$ is normal force; $F_{f}$ is friction

$$m a = m g \sin \theta - \mu m g \cos \theta$$ $$a = g (\sin \theta - \mu \cos \theta) = 9. 8 1 \cdot (\sin 3 7. 6 {}^ {\circ} - 0. 3 4 3 \cdot \cos 3 7. 6 {}^ {\circ}) = 3. 3 2 \mathrm {m / s ^ {2}}$$

Applying $v_{f}^{2} = v_{i}^{2} + 2ad$ gives

$$v _ {f} ^ {2} = 0 + 2 \cdot 3. 3 2 \cdot 2. 0 0 = 1 3. 2 8$$ $$v _ {f} = \sqrt {1 3 . 2 8} = 3. 6 4 \mathrm {m / s}$$

Answer: e. $3.64\mathrm{m / sec}$

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