Question

A frictionless plane is 10.0 m long and inclined at 41.00. A sled starts at the bottom with an initial speed of 5.50 m/s up the incline. When the sled reaches the point at which it momentarily stops, a second sled is released from the top of the incline with an initial speed v.. Both sleds reach the bottom of the incline at the same moment. (a) Determine the distance that the first sled traveled up the incline. (b) Determine the initial speed of the second sled. m/s

Answer 1

(a) Initial speed of the first sled = 5.5 m/s

So, kinetic energy of this sled,

KE = (1/2) mv^2

Opposing the motion of the sled is the component of gravity which is parallel to the plane which is equal to

F = mgsinθ

One more formula is needed, and that is

W=Fs

or work done equals force x distance

Now W = KE = (1/2) mv^2

Rearrange the formula so distance, s, is the subject

s = W/F

= ((1/2) mv^2)/mgsinθ

= ((1/2)v^2)/gsinθ

Now, given that -

v = 5.5 m/s

g = 9.8 m/s^2

θ = 41 degrees

Therefore,

s = ((1/2)5.5^2)/9.8sin41

=15.125 / 6.43

= 2.35 m

Therefore, distance traveled by the first sled up the incline = 2.35 m (Answer)

(b) Now we find the time it takes for the sled to get to that point, since that is also the same time it takes to go from 2.35 m up the slope back to ground level.

Using s = ut + (1/2)at^2

Where s = 2.35 m, u = 5.5 m/s, a = -9.8sin41 = -6.43 m/s^2

[Please note the negative sign because acceleration opposes motion]

Put the values in the above expression -

2.35 = 5.5t + (1/2) * -6.43 t^2

=> 3.21t^2 - 5.5t + 2.35 = 0

So,

t = [5.5 + sqrt{5.5^2 - 4*3.21*2.35}] / (2*3.21)

= [5.5 + sqrt{5.5^2 - 4*3.21*2.35}] / (2*3.21)

= [5.5 + 0.28] / 6.42 = 0.90 s

Now that is the time that the second sled has to travel from the top to the bottom

Also using s = ut + (1/2)at^2

where s =10 m, t = 0.90 s, a = 6.43 m/s^2

[positive because acceleration increases motion]

Therefore,

10 = u*0.90 + (1/2)*6.43 *(0.90)^2

=> 10 = u*0.90 + 2.60

=> u = (10 - 2.60) / 0.90 = 8.22 m/s

Therefore, initial speed of the second sled = 8.22 m/s (Answer)