1. A 2.0 kg block is released from rest at the top of 22 degree frictionless inc
ID: 1395259 • Letter: 1
Question
1. A 2.0 kg block is released from rest at the top of 22 degree frictionless inclined plane of height 0.65 m. At the bottom of the inclined plane, it collides with and sticks to a block of mass 3.5 kg. The two blocks slide a distance of 0.57 m a horizontal plane before coming to a rest. What is the coefficient of friction of the horizontal surface? 2. A solid cylinder of mass M and radius R starts from rest and rolls without slipping down an inclined plane of length L and height h. Find the speed of its center of mass when the cylinder reaches the bottom of the inclined plane. 3. Suppose that the Sun runs out of nuclear fuel and suddenly collapses to form a so-called white dwarf star, with a diameter equal to that of earth. Assuming no what would then be the new rotation period of the Sun, currently is about 25 days? Assume that the Sun and the white dwarf are uniform spheres. 4. A cylinder having a mass of 1.92 kg rotates about its axis of symmetry. Forces are applied as shown in the figure. F1 = 5.88 N, F2 = 4.13 N and F3 = 2.12 N. Also R1 = 4.93 cm and R2 = 11.8cm. Find the magnitude and direction of the angular acceleration of the cylinder.Explanation / Answer
1)
at the bottom of the plane , let the speed of block is v
Now, using conservation of energy
0.5 * 2 * v^2 = 2 * 9.8 * 0.65
v = 3.6 m/s
Now, let the speed after collision is v1
Using conservation of momentum ,
v1 *(3.5 + 2) = 3.6 * 2
v1 = 1.3 m/s
Now, let the coefficient of friction is u
Now, using work energy theorum
u * (3.5 + 2)* 9.8 * 0.57 = 0.5 * (3.5 + 2 )*1.3^2
u = 0.151
the coefficient of kientic friction is 0.151
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