A 10 kg small block is pulled in the vertical plane along a frictionle

1.	A 10 kg small block is pulled in the vertical plane along a frictionless surfacin the form of an arc of a circle of radius 10 m. The applied force is of 200as shown in the figure. If the block started from rest at A, the speed atwould be: (g = 10 m/s2)(A) V3 m/s(B) 10/3 m/s(C) 100/3 m/s(D) None of these
Answer» Answer: The correct answer will be option D. As the answer is 10√3 m/s Explanation: According to the problem the mass of the BLOCK is 10 kg which is pulled in a vertical plane wit a force ALONG an arc. According to the figure the initial position of the block is A and we re applying F force to pull it to B The block is moving as the force is PASSING energy to the block which is equal to its work done W = force X displacement Now displacement means the shortest PATH. Now from As the two arm has the same length of R Therefore Therefore the displacement will be R Therefore the work done at point A W = 200 x 10 = 2000 Joule Now it is the kinetic energy Then, 2000 = 1/2 mv^2 + (mg(R-Rcos60°) => 2000 = 1/2 10 x v^2 + (10 x 10 (10(1-1/2))) => v = 10√3 m/s

A 10 kg small block is pulled in the vertical plane along a frictionless surfacin the form of an arc of a circle of radius 10 m. The applied force is of 200as shown in the figure. If the block started from rest at A, the speed atwould be: (g = 10 m/s2)(A) V3 m/s(B) 10/3 m/s(C) 100/3 m/s(D) None of these

Answer»

Answer:

The correct answer will be option D. As the answer is 10√3 m/s

Explanation:

According to the problem the mass of the BLOCK is 10 kg which is pulled in a vertical plane wit a force ALONG an arc.

According to the figure the initial position of the block is A and we re applying F force to pull it to B

The block is moving as the force is PASSING energy to the block which is equal to its work done

W = force X displacement

Now displacement means the shortest PATH. Now from

As the two arm has the same length of R

Therefore

Therefore the displacement will be R

Therefore the work done at point A

W = 200 x 10 = 2000 Joule

Now it is the kinetic energy

Then,

2000 = 1/2 mv^2 + (mg(R-Rcos60°)

=> 2000 = 1/2 10 x v^2 + (10 x 10 (10(1-1/2)))

=> v = 10√3 m/s