Below are pictures of two systems (system A-left and system B-right). In system

Question

Below are pictures of two systems (system A-left and system B-right). In system A on the left, a 1kg block lies on a frictionless surface and is accelerating to the right because it is attached to another 1kg block that was released from rest and allowed to fall due to gravity. System B on the right shows a 1kg block on a frictionless ramp accelerating to right because it is being pulled by a force of 9.8N. a. Sum the vertical and horizontal forces on top block for only system A below. sigmaF_x: sigmaF_y: b. Sum the vertical and horizontal forces on the hanging block in system A below. sigmaF_X: sigmaF_y: c. Find the accelerations of system A and system B. Acceleration system A = m/Acceleration System B = m

Explanation / Answer

Let us assume that the tension in the string for the system A is T Newtons and the acceleration to be a m/s2

a.) For the top block in the system A, we can write for the forces along the horizontal to get:

Sum of force along horizontal = Fx =  T newtons

For the vertical direction we can write:

Fy = N - mg

b.) For the hanging block:

Forces along the vertical direction = Fy = T- Mg

There are no forces acting on the hanging block along the horizontal direction, hence we get:

Fx = 0

c.) For system A, we have:

Tension along the horizontal = mass x acceleration

or, T = a

and g - T = a

or, a = g/2 = 4.9 m/s2

For system B:

Tension in the string = Ma

or, 9.8 = a

Therefore acceleration of system B = 9.8 m/s2

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