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2 years ago

6

Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0 incline. The coeffi cient of kinetic friction between the bl

ock and the incline is mk 0.436. Determine (a) the work done by the force of gravity, (b) the work done by the friction force between block and incline, and (c) the work done by the normal force. (d) Qualitatively, how would the answers change if a shorter ramp at a steeper angle were used to span the same vertical height

1 answer:

UkoKoshka [18]2 years ago

5 0

Answer:

a) $W_g =61.25J$

b) $W_k = -46.25J$

c) $W_N = 0$

d) $W_g$ would be the same.

$W_k$ would decrease.

$W_N$ would be the same.

Explanation:

a) On an inclined plane the force of gravity is the sine component of the weight of the block.

$F_g = mg\sin(\theta) = 5(9.8)\sin(30^\circ)\\W_g = F_g x = 5(9.8)\sin(30^\circ)2.5 = 61.25J$

b) The friction force is equal to the normal force times coefficient of friction.

$F_k = -mg\cos(\theta)\mu_k = -5(9.8)cos(30^\circ)0.436 = -18.5 N\\W_k = -F_kx = -46.25J$

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.

d) The relation between the vertical height and the distance on the ramp is

$h = x\sin(\theta)$

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of $x\sin(\theta)$ .

The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.

The work done by the normal force would still be zero.

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3 years ago

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Answer:

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Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

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Answer:

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