Question

A block of mass m1 = 290 g is at rest on a plane that makes an angle θ = 30° above the horizontal. The coefficient of kinetic friction between the block and the plane is μk = 0.10. The block is attached to a second block of mass m2 = 200 g that hangs freely by a string that passes over a frictionless and massless pulley. Find its speed when the second block has fallen 30.0 cm.

Answer 1

To find the speed of the second block, we can use the principle of conservation of energy. Assuming the system is frictionless, we equate the initial potential energy of the second block to its final kinetic energy, and solve for the final speed.

Step-by-step explanation:

To find the speed of the second block, we can use the principle of conservation of energy. As the second block falls, it gains gravitational potential energy which gets converted into kinetic energy. Assuming the system is frictionless, we can equate the initial potential energy of the second block to its final kinetic energy:

m2 * g * h = 0.5 * m2 * v^2

where m2 is the mass of the second block, g is the acceleration due to gravity, h is the distance the second block has fallen, and v is the final speed of the second block.

With the given values, we have m2 = 0.2 kg, g = 9.8 m/s^2, and h = 0.3 m, so we can solve for v:

0.2 * 9.8 * 0.3 = 0.5 * 0.2 * v^2

Simplifying the equation, we get v^2 = 0.2 * 9.8 * 0.3 * 2 / 0.2 = 0.882 m^2/s^2. Taking the square root, we find that the speed of the second block is approximately 0.94 m/s.