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3 years ago

If the mass of the block is 5 kg and the speed 7 m/s, what is the work done on the block?

Physics

1 answer:

Kamila [148]3 years ago

8 0

Answer:

A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms

2 ).

Mass of the block=5kg

Coeffecient of friction=0.2

external applied force, F=40N

The angle at which the force is applied=30degree

So the horizontal component of force=Fcos30=40×

23 =20 3 N

While the uertical component of the force acting in upward direction=Fsin30=40× 21

=20N

The normal reaction from the surface (N)=mg−Fsin30=50−20=30N

So the ualue of limiting friction=μN=0.2×30=6N

Hence the net horizontal force on the block=Fcos30=μN=20

N−6N=28.64N

The horizontal acceleration of the block=

Fcos30−μN = 528.64

=5.73m/s 2

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Two people carry a heavy electric motor by placing it on a light board 2.45 m long. One person lifts at one end with a force of

maxonik [38]

Answer:

W=1055N

Explanation:

In order to solve this problem, we must first do a drawing of the situation so we can visualize theh problem better. (See attached picture)

In this problem, we will ignore the board's weight. As we can see in the free body diagram of the board, there are only three forces acting on the system and we can say the system is in vertical equilibrium, so from this we can say that:

$\sum F=0$

so we can do the sum now:

$F_{1}+F_{2}-W=0$

when solving for the Weight W, we get:

$W=F_{1}+F_{2}$

and now we can substitute the given data, so we get:

W=410N+645N

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A coach is hitting pop flies to the outfielders. if the baseball (m= 145 g) stays in contact with the bat for 0.04 s and leaves

vekshin1

At the start, the ball is at rest and therefore, u=0 m/s. As it leaves the bat, v= 50 m/s

From equations of motion, v=u+at = at (since u=o)
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