A block of mass m = 2.00 kg is released from rest at the top of an inclined plan

Question

A block of mass m = 2.00 kg is released from rest at the top of an inclined plane as seen in the figure. The block starts out at height h = 0.100 m above the top of the table, the table height is H-2.00 m, and 0 41.0° (a) What is the acceleration (in m/s) of the block while it slides down the incline? m/s (b) What is the speed (in m/s) of the block when it leaves the incline? m/s (c) At what horizontal distance (in m) from the end of the table will the block hit the ground? (d) How long (in s) from when the block is released does it hit the ground? (e) Does the block's mass affect any of your above answers? Yes NO

Explanation / Answer

a)

Acceleration of the block

a= g sin x= 9.8* sin 41= 6.43 m/s^2

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b)

Velocity before fall

u= sqrt(2gh)= 1.4 m/s

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c)

Time of flight of block

2= 1.4 sin 41*t + 0.5* 9.8t^2

t= 0.552 s

Horizontal range, R= 1.4 cos 41* 0.552= 0.583 m

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d) t= 0.552 seconds

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e) no, all the calculations are independent of mass of the vlobl.

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VommCom in case any doubt.. good luck

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