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Musya8 [376]
3 years ago
11

A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.

How much force is the person currently exerting?
Physics
1 answer:
alexdok [17]3 years ago
7 0

Here when an object is placed on the level floor then in that case there are two forces on the object

1). Weight of object downwards (mg)

2). Normal force due to floor which will counterbalance the weight (N)

so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

F_g = F_n + F_a

given that

F_g = mg

here

m = 30 kg and

g = acceleration due to gravity = 10 m/s^2

F_n = 150 N

now from above equation

30*10 = 150 + F_a

F_a = 300 - 150 = 150 N

So force applied by the person must be 150 N

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Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

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A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo
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Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

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Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds

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Answer:

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Q = - 2.25 x 10^4 J

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