A 50.0-kg block is being pulled up a 13.0° slope by a force of 250 N which is parallel to the slope, but the block does not slid
1 answer:
<span>Weight of block, Wb = mass*gravity = 50*9.8 = 490 N</span>
Since block is being pulled up by a 13-degree slope
Therefore, Force which is acting parallel to the slop:
<span> F p =490 Sin </span>= 110.2N
Force which is acting perpendicular to the slope:
<span> Fv =490 Cos</span> = 477.4 N
Net force can be given as follows:
<span>F n = (250 - 110.2 - 0.2*</span>477.4) N
<span>Fn=44.3N</span>
Now acceleration is given by the ratio of force to mass
<span>a = Fn/m</span>
<span>=44.3/50 = 0.89 ms^<span>-2</span></span>
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Answer:
the magnitude of the work done by the two blocks is the same.
Explanation:
The work done by block a on block b is given by:
where Fa is the force exerted by block a on block b, and d is the distance they cover.
The work done by block b on block a is given by:
where Fb is the force exerted by block b on block a, and d is still the distance they cover.
For Newton's third law, the force exerted by block a on block b is equal to the force exerted by block b on block a, therefore
and so