1. 1.2 m 650 N/m 0.80 m Target Note: Figure not drawn to scale. Block A of mass

Question

1. 1.2 m 650 N/m 0.80 m Target Note: Figure not drawn to scale. Block A of mass 4.0 kg is on a horizontal, frictionless tabletop and is placed against a spring of negligible mass and spring constant 650 N/m. The other end of the spring is attached to a wall. The block is pushed toward the wall until the spring has been compressed a distance x, as shown above. The block is released and follows the trajectory shown, falling 0.80 m vertically and striking a target on the floor that is a horizontal distance of 1.2 m from the edge of the table. Air resistance is negligible (a) Calculate the time elapsed from the instant block A leaves the table till it strikes the floor 2(0 9.8lmis1 (b) Calculate the speed of the block as it leaves the table t o.40S (c) Calculate the distance x the spring was compressed | 6So 2 X0.24 m

Explanation / Answer

Given

spring constant k = 650 N/m

compression distance x = 0.24 m

the elastic potential energy is

U = 1/2 * k * x^2

U = 1/2 * 650 * (0.24)^2

U = 18.72 J (This is elastic potential energy)

The kinetic energy is

K = 1/2 * m * (v0x)^2

k = 1/2 * 4 * 3^2

k = 18 J-----------------------------------Answer

b)

The initial horizontal velocity when it leaves the table is

v0x = 3 m/s (this is same when it hits the ground)

vx = v0x = 3 m/s

vertical component v0y = 0

the vertical component when it hits the ground is

vy = v0y - (g*t)

vy = 0 - (9.8 * 0.4)

vy = - 3.92 m/s

the total velocity when it hits the ground is

v = sqrt((vx)^2 + (vy)^2)

v = sqrt(3^2 + (-3.92)^2)

v = 4.93 m/s-----------------------------------Answer

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