Question

Physics help ??

1. An applied force of 109 N is used to accelerate an object to the right across a frictional surface. The object encounters 19.4 N of friction. The object weighs 88.5 N. Calculate the (a) normal force, (b) the net force, (,c) coeficcient of friction (mu, µ), (d) the mass and (e) the acceleration.

Answer 1

See below.

Explanation:

We are given:

• `abs(vecF_"applied")=109"N"`

• `"weight" = 88.5"N"`

• `abs(vecf)=19.4"N"`

a.) Find the normal force.

Let's start with a force diagram.

where `vecn` is the normal force, `vecf_k` is the force of kinetic friction, `vecF_a` is the applied force, and `vecF_g` is the force of gravity.

Let's define to the left and downward as negative directions. We can then write statements of the net force as follows:

`F_"net, x"=sumF_x=F_a-f_k=ma_x`

`F_"net, y"= sumF_y=n-F_g=ma_y`

Since the object does not accelerate vertically (does not move up or down relative to the horizontal), we have:

`n-F_g=0`

`=>n=F_g`

Since `F_g=mg`, this ultimately provides `n=mg`. In other words, the normal force is equal and opposite (in direction) to the gravitational force acting on the object.

We are given the object's weight, which in this case is the gravitational force, and so the normal force is given by `color(blue)(n=88.5"N")`.

b.) Find the net force.

We can refer to our net force statements from part a.

We know that for the net force parallel we have `F_a-f_k` and for the net force perpendicular we have `0` (equilibrium for this direction).

Then `F_"net, x"=F_"net"=109"N"-19.4"N"=color(blue)(89.6"N")`

c.) Find the coefficient of friction.

We know that the force of friction is given by `vecf=muvecn` where `mu` is the coefficient of friction, and `vecn` is the normal force. We are given the magnitude of the fictional force and we have already found the magnitude of the normal force, so we have:

`mu=f/n`

`=>mu=19.4/88.5`

`=>color(blue)(mu=0.219)`

d.) Find the mass.

As discussed above, we are given the weight of the object, which is equivalent to the force of gravity acting on the object. Therefore we know that `88.5=mg`

`=>m=(88.5"N")/g`

`=>m=(88.5"N")/(9.81"m"//"s"^2)`

`=>color(blue)(m~~9.02"kg")`

e.) Find the acceleration.

We have established that there is no acceleration in the y-direction. Therefore, all acceleration experienced by the object is parallel to the motion (x-direction). We can find the acceleration using Newton's second law.

We found above that `F_a-f_k=ma_x`

We found the mass in part b and the net force in part b, so we have:

`=>a_x=(F_a-f_k)/(m)`

`=>a_x=(89.6"N")/(9.02"kg")`

`=>color(blue)(a=9.83"m"//"s"^2)`

Answer 2

Frictional force (`f`) acting at the interface of two objects is given as `mu*N` (where, `N` is the normal force and `mu` is coefficient of friction)

Here, `N=mg=88.5N` (given) or, `m=N/g =8.85 Kg`(`m` is the massof the object)

And, `f=19.4 N`

So,`mu =f/N=0.22`

Net force(`F`) is acting along the direction of movement i.e `(109-19.4)N=89.6N`

So,acceleration of the object is `F/m=89.6/8.85=10.13 ms^-2`