Question

a block of mass M equal to 9 kg is moving with uniform velocity of 2 metre per second on a horizontal surface under the action of a constant force acts before acceptance angle 37 degree to the horizontal find the work done by the force during the interval of 4 seconds of motion​

Answer 1

Answer:

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Answer 2

Concept Introduction:-

A huge body of substance with no visible shape is known to as mass.

Given Information:-

We have been given that a block of mass M equal to $9$ kg is moving with uniform velocity of $2$ metre per second on a horizontal surface under the action of a constant force acts before acceptance angle $37$ degree to the horizontal.

To Find:-

We have to find that work done by the force during the interval of $4$ seconds of motion​

Solution:-

According to the problem

Now since there is no acceleration on the block in the horizontal or the vertical direction, the sum of the forces in the horizontal direction and the vertical direction balance each other.

Now in the vertical direction, we can see there is the sine component of the force F in the downward direction and the weight of the body mg acting in the downward direction. The normal reaction force on the body due to the surface is in the upward direction.

So we can set the equation as,

$N=Fsin37^\circ+mg$

Now in the question we are given $m=9$ kg and the acceleration due to gravity is taken approximately $g=10m/s^2$. The value of $sin37^\circ$ can be approximately taken as, ⇒$sin37^\circ=\frac{3}{5}$

So substituting these values in the equation we get,

⇒$N=\frac{3}{5}F+9\times 10$

Therefore the normal reaction force on the body is equal to

⇒$N=\frac{3}{5} F+90$

Now in the horizontal direction, we can see that the cosine component of the applied force is balanced by the force of friction. The force of friction is denoted here by f and it can be calculated by $f=$μ$N$ where μ is the coefficient of friction between the $2$ surfaces. In the diagram in the question, we are given the coefficient of friction as, μ$=\frac{1}{3}$ and the normal reaction force is calculated above as,

$N=\frac{3}{5} F+90$. Substituting these values we get the force of friction as,

⇒f$f=\frac{1}{3}(\frac{3F}{5}+90)$

On opening the bracket we get,

⇒$f=\frac{F}{5}+30$

Therefore in the horizontal direction we can write the equation of motion of the body as,

⇒$f=Fcos37^\circ$

Substituting values,

⇒$\frac{F}{5}+30=Fcos37^\circ$

Now the value of $cos37^\circ$ is approximately, $cos37^\circ=45$

Hence we get in the equation,

⇒$\frac{F}{5}+30= \frac{4F}{5}$

On taking $\frac{F}{5}$ to the other side we have,

⇒$\frac{4F}{5}-\frac{F}{5}=30$

On doing the subtraction we get,

$\frac{4F}{5}-\frac{F}{5}=30\\\frac{3F}{5}=30\\F=\frac{30\times 5}{3}\\=50N$

This is the force exerted on the body. Now the displacement of the body is given by the product of velocity and time since the acceleration is zero. Therefore, we have

⇒$S=v\times t$

According to the problem, the velocity is $v=2$m/s and time is $t=4$s. On doing the product we get

⇒$S=2\times 4=8$m

Now the work done on the body is given by the formula,

⇒$W=FS =FScos$ θ

Where we have $F=50$N, $S=8$m and the angle between the force and displacement as, θ$=37^\circ$

Hence substituting we get,

⇒$W=50\times 8\times cos37^\circ$

Since $cos37^\circ=45$,

⇒$W=400\times 45$

Cancelling we get, $W=80\times 4=320$J.

Final Answer:-

The work done by the force is $320$ J.

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