a block is given a brief push up a 20 incline to give it an initial speed of 6.0

Question

a block is given a brief push up a 20 incline to give it an initial speed of 6.0 m/s. the block slides up the incline and returns to the initial position. If the coefficient of kinetic friction between the block and the incline is 0.25.
A. How far up the incline does the block slide? B. How much time doesnit take for the block to reach this top position? C. How much time does it take for the block to slide from this top postion back to its starting position? a block is given a brief push up a 20 incline to give it an initial speed of 6.0 m/s. the block slides up the incline and returns to the initial position. If the coefficient of kinetic friction between the block and the incline is 0.25.
A. How far up the incline does the block slide? B. How much time doesnit take for the block to reach this top position? C. How much time does it take for the block to slide from this top postion back to its starting position? a block is given a brief push up a 20 incline to give it an initial speed of 6.0 m/s. the block slides up the incline and returns to the initial position. If the coefficient of kinetic friction between the block and the incline is 0.25.
A. How far up the incline does the block slide? B. How much time doesnit take for the block to reach this top position? C. How much time does it take for the block to slide from this top postion back to its starting position?

Explanation / Answer

= 20o
vi = 6.0 m/s
uk = 0.25

(A)
Let the distance moved along the incline be d.
Let the max height gained be h.

sin(20) = h/d
h = d*sin(20)

Using Energy conservation,
Initial Kinetic Energy = Potential Energy at max Height + Energy lost in Friction
1/2 * m * v^2 = m*g*h + uk*m*g*cos(20) * d
1/2 * v^2 = g*d*sin(20) + uk*g*cos(20) * d
1/2 * 6.0^2 = 9.8*d*sin(20) + 0.25*9.8*cos(20) * d
d = 3.18 m

(B)
vf^2 = vi^2 + 2*a*s
0 = 6.0^2 - 2*a*3.18
a = 5.66 m/s^2

vf = vi + a*t
0 = 6.0 - 5.66*t
t =  1.06 s
Time does it take for the block to reach this top position, t =  1.06 s

(C)
d = 3.18

m*g*sin(20) - uk* m*g*cos(20) = m*a
9.8*sin(20) - 0.25*9.8*cos(20) = a
a = 1.05 m/s^2

s = u*t + 1/2*a*t^2
3.18 = 0 + 1/2*1.05*t^2
t = 2.46 s

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