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2 years ago

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A 46.0-kg box is being pushed a distance of 8.80 m across the floor by a force P whose magnitude is 171 N. The force P is parall

el to the displacement of the box. The coefficient of kinetic friction is 0.250. Determine the work done on the box by (a) the applied force, (b) the friction force, (c) the normal force, and (d) by the force of gravity. Be sure to include the proper plus or minus sign for the work done by each force.

1 answer:

dolphi86 [110]2 years ago

8 0

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

$W_P=Px$

P: applied force = 171N

x: distance in which the for P is applied = 8.80m

you replace the values of P and x and obtain:

$W_P=(171N)(8.80m)=1504.8J$

(b) The work don by the friction force is:

$W_f=F_fx=\mu N x=\mu Mg x$

μ = coefficient of kinetic friction = 0.250

M: mass of the box = 46.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J$

(c) The Normal force is

$N=Mg=(46.0kg)(9,8m/s^2)=450.8N$

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

$W_N=0J$

(d) the same as before:

$W_g=0J$

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In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor

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(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

$W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}$ .....(I)

Where, $F(x)$ = net force

We know, the net force is the sum of forces.

So, $\sum{F}=ma$

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

$F(x)+(-F(f))=ma$

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

$W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

We need to find the value of $\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

Using newton's second law

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}$ ...(II)

We know that,

Acceleration is rate of change of velocity.

$a=\dfrac{dv}{dt}$

Put the value of a in equation (II)

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

Now, the work done by the net force on the block is,

$W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

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