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dybincka [34]
3 years ago
12

A block of mass 5 kg is placed on a rough table having a coefficient of static friction μs = 0.2.What is the force required to m

ove the block?
Physics
1 answer:
just olya [345]3 years ago
5 0

Answer:

<em>We need to (at least) apply a force of 9.8 N to move the block</em>

Explanation:

<u>Second Newton's Law</u>

If a net force F_n different from zero is applied to an object of mass m, then it will move at an acceleration a, given by

F_n=ma

If we apply a force F to an object placed on a rough surface, the only way to make it move is to beat the friction force  which is given by

F_r=\mu F_N

Where \mu is the static friction coefficient and F_N is the normal force exerted by the table to the object. Since there is no motion in the vertical direction the normal force equals the weight of the object:

F_N=mg=5\ kg\ 9.8\ m/s^2=49\ N

The friction force is

F_r=0.2 (49)=9.8\ N

Thus, we need to (at least) apply a force of 9.8 N to move the block

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It's C

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Natasha_Volkova [10]

The hydrogen fusion process will begin after the protostar reaches a temperature of 10 million degrees kelvin, and it will then turn into a stable star.

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3 0
1 year ago
A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.
Ede4ka [16]

(a) The work required to get the coaster to the top of the first hill is  1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

P = Fd/t

P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

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6 0
2 years ago
A girl throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounc
erma4kov [3.2K]

Answer:

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Applying the impulse-momentum equation;

Impulse = change in momentum

Ft = m∆v

F = (m∆v)/t

Where;

F = force

t = time

m = mass

∆v = v2 - v1 = change in velocity

Given;

m = 0.80 kg

t = 0.050 s

The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.

v2 = 25 m/s

v1 = -25 m/s

∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s

Substituting the values;

F = (m∆v)/t

F = (0.80×50)/0.05

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

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2 years ago
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