A frictionless plane is 10.0 m long and inclined at 36.0°. A sled starts at the

Question

A frictionless plane is 10.0 m long and inclined at 36.0°. A sled starts at the bottom with an initial speed of 4.20 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 vi. Both sleds reach the bottom of the incline at the same moment.

(a) Determine the distance that the first sled traveled up the incline.
m

(b) Determine the initial speed of the second sled.
m/s

Please show a detailed solution

Explanation / Answer

acceleration acting on first sled,
a= g sin (theta),
theta = degree of inclination
a = -9.81 * sin 36 = -5.77 m/s^2

time taken by sled to stop, ie v = 0 m/s
v = u + at
u = initial velocity
0 = 4.2 + (-5.77 )* t
t = 0.73 s
   now the distance travelled in these 0.73 s ,
   S = ut + 0.5 at^2
   S = 4.2 *0.73 + 0.5 *(-5.77) 0.73^2 = 1.53 m
   ie sled traveled 1.53 m up the inclined plane.

   first sled takes same time to get back to bottom again. ie 0.73 s
    for second sled to reach the bottom at same time as first sled, it has to be given some initial velocity
  
   S = ut +0.5 *at^2
   1.53 = u*0.73   + 0.5*(-5.77)*0.73^2
   u = 4.2 m/s
  
   hence initil velocity of second sled will be 4.2 m/s
  
   (Inside info: note that both sled have been given the same initial speed, this is so bcoz they are of equal masses in this particular example hence they have same motions)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.