The 20 kg block is sliding down an inclined plane at a 30 ° inclined friction and is stopped by a spring with a spring constant k = 2 x 10 ^ 4 N / m. The block shifted to a total distance of 4 m from the point where it was released to the point where it stopped with the resistance of the spring. How many meters is the bow compressed when the block stops? (g = 10 m / s ^ 2)

5:29 PM Sun Nov 5 Tt S An ideal spring is attached to the ceiling. While the spring is held at its relaxed length, a wooden block (M 850 g) is attached to the bottom of the spring, and a ball of clay (m 210g) is pressed onto the bottom of the block so that they stick together. The block+clay are gently lowered through a distance of d 2.7 cm and are then released, at which point they hang motionlessly from the bottom of the - spring. After a few minutes have passed, the clay unsticks it- self from the block and falls from rest to the ground. What is the resulting period T of the block's oscillation after the sep- aration occurs? For the limit check, investigate what happens to T if the block+clay only need to be lowered through an extremely short distance (d → 0) before they can be released motionlessly. = ✪ 4% =

A spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is 0.90 kg, and the spring has a spring constant of 53 N/m. The coefficient of static friction between the box and the table on which it rests is μs = 0.77. How far can the spring be stretched from its unstrained position without the box moving when it is released?

This relates the initial speed of the pendulum box to the height that it goes: V =squr 2 g h .  The numbers: g = 9.81 m/s2 h = 3.6 cm V = squr 2 × 9.81 × 3.6 = 8.4  Correct the calculation or the answer