- #1
musicfairy
- 101
- 0
Can someone check if I did this problem right?
A horizontal force F is applied to a small block of mass m1 to make It slide along the top of a larger block of mass m2 and length L. The coefficient of friction between the blocks is [tex]\mu[/tex]. The larger block slides without friction along a horizontal surface. The blocks start from rest with the small block at one end of the larger block.
a) Find the acceleration of each block. a1 and a2, relative to the horizontal surface.
b) In terms of L, a1, and a2, find the time t needed for the small block to slide off the end of the larger block.
This is what I did. f is friction, F is the applied force
a)
f = F - m1a1
f = m2a2
f = [tex]\mu[/tex]m1g
[tex]\mu[/tex]m1g = F - m1a1
a1 = (F - [tex]\mu[/tex]m1g) / m1
a2 = [tex]\mu[/tex]m1g/m2
b)
L = .5(a1 - a2)t2
t = [tex]\sqrt{2L / (a1 - a2)}[/tex]
A horizontal force F is applied to a small block of mass m1 to make It slide along the top of a larger block of mass m2 and length L. The coefficient of friction between the blocks is [tex]\mu[/tex]. The larger block slides without friction along a horizontal surface. The blocks start from rest with the small block at one end of the larger block.
a) Find the acceleration of each block. a1 and a2, relative to the horizontal surface.
b) In terms of L, a1, and a2, find the time t needed for the small block to slide off the end of the larger block.
This is what I did. f is friction, F is the applied force
a)
f = F - m1a1
f = m2a2
f = [tex]\mu[/tex]m1g
[tex]\mu[/tex]m1g = F - m1a1
a1 = (F - [tex]\mu[/tex]m1g) / m1
a2 = [tex]\mu[/tex]m1g/m2
b)
L = .5(a1 - a2)t2
t = [tex]\sqrt{2L / (a1 - a2)}[/tex]
Last edited: