A block of mass 40 g is attached with a spring. If the normal length of spring (without attaching block) was 10 cm and after attaching the block, new length of spring is 12 cm. Calculate the spring constant as per the Hook’s law.
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- answer part bA crate of mass m1 slides down a well-lubricated hill of height h, with negligible friction. At the bottom, where it is moving horizontally, it collides with another crate, of mass m2, that initially was sitting at rest and that is attached to a wall by a spring of spring constant k that initially is at its equilibrium length. Assume that the spring itself has negligible mass. a)Given that the distance d that the crates compress the spring is d=0.35 m, calculate the speed v2 of the crates immediately after the collision, in units of meters per second. Use the following values:k=950 N/mm1=2.4 kgm2=2.6 kgμ=0.49g=9.8 m/s2 b)What was the speed of the crate of mass m1 just before the collision with the second block, in meters per second? c) What is the height h of the hill, in meters?A crate of mass m1 slides down a well-lubricated hill of height h, with negligible friction. At the bottom, where it is moving horizontally, it collides with another crate, of mass m2, that initially was sitting at rest and that is attached to a wall by a spring of spring constant k that initially is at its equilibrium length. Assume that the spring itself has negligible mass. a)Given that the distance d that the crates compress the spring is d=0.35 m, calculate the speed v2 of the crates immediately after the collision, in units of meters per second. Use the following values:k=950 N/mm1=2.4 kgm2=2.6 kgμ=0.49g=9.8 m/s2 b) What was the speed of the crate of mass m1 just before the collision with the second block, in meters per second? c)What is the height h of the hill, in meters?
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