In this activity, you will analyze the motion of a cart being pulled up an inclined plane at a constant speed. The angle of the
incline can be modified by 10 degrees increments between the values of 30 degrees and 90 degrees. Three different masses can be selected - 2.0kg, 3.0kg, and 4.0kg in each simulation the cart is pull d to the same height - 1.0 meter above the original starting position. For each simulation, the force that must be applied is reported on the screen. The displacement of the cart can be measured using the cm-ruler that is displayed for each trial. Can some help me
1 answer:
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Answer:
A= 148.92 m/s²
Explanation:
Given that
U(x,y) = (6.00 )x² - (3.75 )y ³
m= 0.04 kg
Now force in the x-direction
Fx= - dU/dx
U(x,y) = (6.00 )x² - (3.75 )y ³
dU/dx= 12 x
When x=0.4 m
dU/dx= 12 x 0.4 = 4.8
So we can say that
Fx= - 4.8 N
From Newtons law
F= m a
- 4.8 = 0.04 x a
a = -120 m/s²
Acceleration in x direction ,a = -120 m/s²
In y -direction
F= - dU/dy
U(x,y) = (6.00 )x² - (3.75 )y ³
dU/dy = 0 - 3.75 x 3 y²
When y = 0.56 m
dU/dy = - 3.75 x 3 x 0.56 x 0.56
dU/dy = - 3.52
So we can say that force in y -direction
F= 3.52 N
F= m a'
3.52 = 0.04 x a'
a'=88.2 m/s²
acceleration in y direction is 88.2 m/s²
The resultant acceleration
A= 148.92 m/s²