In a physics lab experiment, a spring clamped to the table shoots a 22 g ball ho

Question

In a physics lab experiment, a spring clamped to the table shoots a 22 g ball horizontally. When the spring is compressed 22 cm , the ball travels horizontally 4.9 m and lands on the floor 1.4 m below the point at which it left the spring.
What is the spring constant? In a physics lab experiment, a spring clamped to the table shoots a 22 g ball horizontally. When the spring is compressed 22 cm , the ball travels horizontally 4.9 m and lands on the floor 1.4 m below the point at which it left the spring.
What is the spring constant? What is the spring constant?

Explanation / Answer

PE of the spring = KE of the block

0.5kx^2= 0.5mv^2

v = x*sqrt(k/m)

v = 0.22 * sqrt(k/0.022)

v = 1.4832 sqrt k

x = Ux*t

y = uyt - 0.5gt^2

t = x/Ux = 4.9/1.4832 sqrt k

t = 3.3 / sqrt k - --------(1)

1.4 = 0.5*9.8*(3.3/sqrt k)^2

k = 38.2 N/m

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