A 1.0 kg block slides down an inclined plane of theta equals29 0 from the horizontal. If the block starts from rest and hits the bottom in 4.2 s, what is the speed of the block, in the unit m/s, at the bottom of the incline? Assume a frictionless plane.
Answers
Answer:
Below figure shows the forces on the object. The two forces acting on the block are the normal force, n, and the weight, mg. If the block is considered to be a point mass and the x axis is chosen to be parallel to the plane, then the free-body diagram will be as shown in the figure below. The angle θ is the angle of inclination of the plane.
Applying Newton’s second law for the accelerating system (and taking the direction up the plane as the positive x direction), we have
Step-by-step explanation:
hope it helped
(a) Below figure shows the forces on the object. The two forces acting on the block are the normal force, n, and the weight, mg. If the block is considered to be a point mass and the x axis is chosen to be parallel to the plane, then the free-body diagram will be as shown in the figure below. The angle θ is the angle of inclination of the plane.
Applying Newton’s second law for the accelerating system (and taking the direction up the plane as the positive x direction), we have
∑F
y
=n−mgcosθ=0:n=mgcosθ
∑F
x
=−mgsinθ=ma:a=–gsinθ
(b) When θ=15.0
0
,
a=−2.54m/s
2
.
(c) Starting from rest,
v
f
2
=v
i
2
+2a(x
f
−x
i
)=2aΔx
∣v
f
∣=
2∣a∣Δx
=
2∣−2.54m/s
2
∣(2.00m)
=3.19m/s