The system in Figure Q6 slides down an inclined plane. Object 1 is 10 kg whereas

Question

The system in Figure Q6 slides down an inclined plane. Object 1 is 10 kg whereas object 2 is 2 kg. the coefficient of static friction between the surface and m, is 0.6. Calculate the acceleration of the blocks and the tension in the cord, ANS: a = -1.46 m/s^2;T= 16.7 N Determine the tension in the ropes and calculate the minimum mass required that will cause the 300 kg block to just start moving to the right. (Coefficient of static friction between the block and surface is 0.3). ANS: T,_1= 392.4 N; T_2= 1060 N; m_2= 108 kg A ball is thrown straight upward with an angle of 30degree to the horizontal and is caught directly at the top edge of a building. the lop edge is 5.0 m above the throwing point. Calculate the velocity at which it was thrown and the time taken for it to reach the top of the building. ANS: u=19.8 m/s; t=1.01 s. A 7500kg truck travelling at 5 0ms east collides with a 15(H) kg car moving at 20 m/s in a direction 30degree south from cast Alter collision, the two vehicles remain tangled together. Name the collision type and calculate the velocity and in which direction does the wreckage begin to move.

Explanation / Answer

8) Let initial velocity = v0

so first split up v0 into it's horizontal and vertical components. In this case, horizontal is v0cos30, and vertical is v0sin30.
When the ball reaches it's maximum height, it has vertical velocity of zero.

So, using the equation v^2 = u^2 + 2as, and replacing v=0, u=v0sin30, a = -9.8, s = 5, and then solving for v0, we find that v0 = 19.8 m/s.

Time = v0sin30/9.8 = 19.8*0.5/9.8 = 1.01 s

9) force = rate of change of momentum = m(dv/dt)

F = 2 * 6/(7*10^-4) = 1.7*10^4 N

10) M1 V1 + M2 V2 = (M1 + M2) Vf (vectors)
Vf = (M1 V1 + M2 V2)/(M1 + M2)
Assume the coordinate system: X ---> East ; Y---> South
then:
Vfx = (M1 V1 - M2 V2 sin30°)/(M1 + M2) = 2.5 m/s
Vfy = M2 V2 cos30°/(M1+M2) = 2.89 m/s
|Vf| = 3.15 m/s

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