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maniacp08
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A 6.0kg block slides 1.5m down a frictionless incline that makes an angle of 60 degrees with the horizontal.
a) Find the work done by each force when the block slides 1.5m(measured along the incline)
b) What is the total work done on the block?
c) What is the speed of the block after it has slid 1.5m, if it starts from rest?
d) What is its speed after 1.5, if it starts from initial speed of 2.0m/s?
I drew a free body diagram and there are total of 3 forces.
Normal Force = 58.86 cos 60 = 29.43N
Force of gravity = 6.0 * -9.81 = -58.86N
Force parallel = 58.86 sin 60 = 50.97N
Work = F cos Theta * displacement
Do I need to incorporate the Cos Theta in the work equation below?
Will the work for these forces be:
Normal force = 29.43N * 1.5
Force of gravity = -58.86N * 1.5
Force Parallel = 50.97N * 1.5
Will the answer for B - the total work done be the sum of the Work from all the forces?
Is the answer for C related to F = MA?
Force parallel = M A
50.97N = 3 * A
A = 16.99 m/s^2
Using Vf^2 = Vi^2 + 2ad
Vf^2 = 0 + 2(16.99)(1.5)
Vf^2 = 50.97
Vf = 7.13 m/s
And for question D, is same kinematics equations but substitute 2.0m/s for initial velocity?
Any help would be great! Thanks.
a) Find the work done by each force when the block slides 1.5m(measured along the incline)
b) What is the total work done on the block?
c) What is the speed of the block after it has slid 1.5m, if it starts from rest?
d) What is its speed after 1.5, if it starts from initial speed of 2.0m/s?
I drew a free body diagram and there are total of 3 forces.
Normal Force = 58.86 cos 60 = 29.43N
Force of gravity = 6.0 * -9.81 = -58.86N
Force parallel = 58.86 sin 60 = 50.97N
Work = F cos Theta * displacement
Do I need to incorporate the Cos Theta in the work equation below?
Will the work for these forces be:
Normal force = 29.43N * 1.5
Force of gravity = -58.86N * 1.5
Force Parallel = 50.97N * 1.5
Will the answer for B - the total work done be the sum of the Work from all the forces?
Is the answer for C related to F = MA?
Force parallel = M A
50.97N = 3 * A
A = 16.99 m/s^2
Using Vf^2 = Vi^2 + 2ad
Vf^2 = 0 + 2(16.99)(1.5)
Vf^2 = 50.97
Vf = 7.13 m/s
And for question D, is same kinematics equations but substitute 2.0m/s for initial velocity?
Any help would be great! Thanks.