Question

A frictionless surface is inclined at an angle of 28.8° to the horizontal. A 270-g block on the ramp is attached to a 75.0-g block using a pulley. a)Find the tension in the string and the magnitude of the acceleration of the 270-g block. b)The 270-g block is released from rest. How long does it take for it to slide a distance of 1.10 m along the surface?

Answer 1

a) As you can see by this diagram, the tension force on the string is equal to T = 75*9.8 which is 735 N.

This means the force that acts on the 270 g block is 735 N in the positive x direction. In the negative x direction the force is the same as -m*g*sin(ø) or -270*9.8*sin(28.8) which evaluates to -1274.72 N. The total force then on the block is -539.72 N. This means that -539.72 = 270*a where a is acceleration. a = -1.999 rounded to the nearest thousandth.

b) If the 270 g block was released from the string then the total force is simply -1274.72 N meaning -1274.72 = 270*a and a = 4.721. If a(t) = 4.721 then v(t) = 4.721t + c. If we let its initial time be 0 and its initial velocity be 0 then 0 = c and v(t) = 4.721t. Its position can then be modeled by p(t) = 2.361*t^2 + c. If we let it's initial position be 0 and it's initial time be 0 then c = 0 and p(t) = 2.361*t^2. we can then plug 1.10 in to the position and solve for t and t is approximately 0.683 s.