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Triss [41]
3 years ago
11

How much work did the movers do (horizontally) pushing a 41.0-kgkg crate 10.6 mm across a rough floor without acceleration, if t

he effective coefficient of friction was 0.60?
Physics
1 answer:
Kitty [74]3 years ago
8 0

Answer:

W=2.60 J

Explanation:

Given that

m = 41 kg

d= 10.6 mm

Coefficient of friction = 0.6

Friction force ,Fr=μ m g

Fr= 0.6 x 41 x 10 N               ( take g= 10 m/s²)

Fr= 246 N

We know that work done by a force given as

W = F . d

Now by putting the values

W= 246 x 10.6 x 10 ⁻³  J

W=2.60 J

Therefore the work done by movers in the horizontal direction will be 2.60 J.The mover does the work against the friction force.

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iogann1982 [59]

Answer:

Explanation:

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\displaystyle \frac{1}{2}m(2v_0)^2+\frac{1}{2}2mv_0^2=\frac{1}{2}mv_1^2+\frac{1}{2}2mv_2^2 \quad (1/m) \quad 6v_0^2=v_1^2+2v_2^2

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4 years ago
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3 years ago
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3 years ago
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6 0
4 years ago
why Galileo's thought experiment about the ball rolling forever on a level surface could only be a thought experiment.
erma4kov [3.2K]
Because in reality there are frictional forces acting on the ball, against the direction of its motion. In fact, because of the friction between the ball and the surface, the ball loses little by little its energy, so its velocity decreases and eventually it stops. 
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