Question

Physics 202 help homework question

Answer 1

a) See free-body diagram in attachment

b) The acceleration down along the ramp is $6.9 m/s^2$

c) The new acceleration is $5.5 m/s^2$

d) The power dissipated by friction is 294.2 W

Explanation:

a)

The free body diagram is shown in attachment (first picture). The x-positive direction is taken as down along the ramp, while the positive y-direction is taken up perpendicular to the ramp.

There are only two forces acting on Prof. Holt:

- Her weight, labelled as (mg) (where m is the mass and g the acceleration of gravity), acting vertically downward

- The normal reaction exerted by the slope on the Prof, labelled with N, acting perpendicularly to the plane in the y-direction

We can further resolve the weight of the Prof along the x- and y- direction, obtaining:

$W_x = mg sin \theta\\W_y = mg cos \theta$

Where $\theta$ is the angle at the base of the ramp.

b)

The equation of the forces along the two directions are:

- x direction:

$W_x = ma$

where a is the acceleration,

- y direction:

$N-W_y = 0$

We can rewrite the two equations as

$mg sin \theta = ma\\N = mg cos \theta$

And from the first equation, we get:

$a=g sin \theta$

Substituting $g=9.8 m/s^2$ and $\theta=45^{\circ}$, we find the acceleration:

$a=(9.8)(sin 45)=6.9 m/s^2$

c)

The new free-body diagram is attached in the second picture. In this case, we also have the frictional force $F_f$, which acts up along the ramp, and whose magnitude is

$F_f = \mu N$

where

$\mu=0.2$ is the coefficient of kinetic friction.

In this situation, therefore, the two equations of motions are:

$N - mg cos \theta = 0\\mg sin \theta - \mu N = ma$

From the first equation we find the normal reaction:

$N=mg cos \theta$

And substituting into the second equation, we can find the acceleration:

$mg sin \theta - \mu mg cos \theta = ma\\a = g sin \theta - \mu g cos \theta = (9.8)(sin 45)-(0.2)(9.8)(cos 45)=5.5 m/s^2$

d)

First of all, we have to calculate the magnitude of the force of friction, which is given by:

$F_f = \mu N = \mu mg cos \theta$

and substituting m = 45 kg and the other data, we find

$F_f = (0.2)(45)(9.8)(cos 45)=62.4 N$

Now we can calculate the work done by the force of friction, which is

$W=F_f d$

where d is the displacement along the ramp, which is given by

$d=\frac{h}{sin \theta}$

where h = 10 m is the height of the ramp. Substituting and evaluating,

$W=-F_f \frac{h}{sin \theta}=-(62.4) \frac{10}{sin(45)}=-882.5 J$

The work done by the friction is negative since friction acts opposite to the direction of motion.

Now, we can finally calculate the power dissipated:

$P=\frac{|W|}{t}$

where

$|W|=882.5 J$ is the magnitude of the work done by friction

t = 3 s is the time interval

Substituting,

$P=\frac{882.5}{3}=294.2 W$

Learn more about slopes, force of friction and power:

brainly.com/question/5884009

brainly.com/question/7956557

#LearnwithBrainly