a) See free-body diagram in attachment
b) The acceleration down along the ramp is [tex]6.9 m/s^2[/tex]
c) The new acceleration is [tex]5.5 m/s^2[/tex]
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:
[tex]W_x = mg sin \theta\\W_y = mg cos \theta[/tex]
Where [tex]\theta[/tex] is the angle at the base of the ramp.
b)
The equation of the forces along the two directions are:
- x direction:
[tex]W_x = ma[/tex]
where a is the acceleration,
- y direction:
[tex]N-W_y = 0[/tex]
We can rewrite the two equations as
[tex]mg sin \theta = ma\\N = mg cos \theta[/tex]
And from the first equation, we get:
[tex]a=g sin \theta[/tex]
Substituting [tex]g=9.8 m/s^2[/tex] and [tex]\theta=45^{\circ}[/tex], we find the acceleration:
[tex]a=(9.8)(sin 45)=6.9 m/s^2[/tex]
c)
The new free-body diagram is attached in the second picture. In this case, we also have the frictional force [tex]F_f[/tex], which acts up along the ramp, and whose magnitude is
[tex]F_f = \mu N[/tex]
where
[tex]\mu=0.2[/tex] is the coefficient of kinetic friction.
In this situation, therefore, the two equations of motions are:
[tex]N - mg cos \theta = 0\\mg sin \theta - \mu N = ma[/tex]
From the first equation we find the normal reaction:
[tex]N=mg cos \theta[/tex]
And substituting into the second equation, we can find the acceleration:
[tex]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[/tex]
d)
First of all, we have to calculate the magnitude of the force of friction, which is given by:
[tex]F_f = \mu N = \mu mg cos \theta[/tex]
and substituting m = 45 kg and the other data, we find
[tex]F_f = (0.2)(45)(9.8)(cos 45)=62.4 N[/tex]
Now we can calculate the work done by the force of friction, which is
[tex]W=F_f d[/tex]
where d is the displacement along the ramp, which is given by
[tex]d=\frac{h}{sin \theta}[/tex]
where h = 10 m is the height of the ramp. Substituting and evaluating,
[tex]W=-F_f \frac{h}{sin \theta}=-(62.4) \frac{10}{sin(45)}=-882.5 J[/tex]
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:
[tex]P=\frac{|W|}{t}[/tex]
where
[tex]|W|=882.5 J[/tex] is the magnitude of the work done by friction
t = 3 s is the time interval
Substituting,
[tex]P=\frac{882.5}{3}=294.2 W[/tex]
Learn more about slopes, force of friction and power:
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