Problem 10.62 In a physics lab experiment, a spring clamped to the table shoots

Question

Problem 10.62 In a physics lab experiment, a spring clamped to the table shoots a 21 g ball horizontally. When the spring is compressed 18 cm , the ball travels horizontally 4.8 m and lands on the floor 1.3 m below the point at which it left the spring. Part A What is the spring constant? Express your answer using two significant figures. k = 5.73   N/m   Problem 10.62 In a physics lab experiment, a spring clamped to the table shoots a 21 g ball horizontally. When the spring is compressed 18 cm , the ball travels horizontally 4.8 m and lands on the floor 1.3 m below the point at which it left the spring. Part A What is the spring constant? Express your answer using two significant figures. k = 5.73   N/m   Problem 10.62 Problem 10.62 In a physics lab experiment, a spring clamped to the table shoots a 21 g ball horizontally. When the spring is compressed 18 cm , the ball travels horizontally 4.8 m and lands on the floor 1.3 m below the point at which it left the spring. In a physics lab experiment, a spring clamped to the table shoots a 21 g ball horizontally. When the spring is compressed 18 cm , the ball travels horizontally 4.8 m and lands on the floor 1.3 m below the point at which it left the spring. In a physics lab experiment, a spring clamped to the table shoots a 21 g ball horizontally. When the spring is compressed 18 cm , the ball travels horizontally 4.8 m and lands on the floor 1.3 m below the point at which it left the spring. Part A What is the spring constant? Express your answer using two significant figures. k = 5.73   N/m   Part A What is the spring constant? Express your answer using two significant figures. k = 5.73   N/m   Part A What is the spring constant? Express your answer using two significant figures. k = 5.73   N/m   k = 5.73   N/m   k = 5.73   N/m   5.73 k = 5.73   N/m  

Explanation / Answer

when the spring compresses 18 cm

KE of the ball = spring energy

0.5 mv^2 = 0.5 kx^2

0.021*v^2 = k*(0.18)^2

v = 1.54 sqrt k

using the equation of motion , we have

x = Ux*t+0.5ax*t^2   

ax= 0

x = Ux *t   

4.8 = 1.54sqrt k * t

t = 4.8 / 1.54*sqrt k

y = Uyt+0.5gt^2

1.3 = 0 + 0.5*9.8*(4.8/1.54)^2 *(1/k)

k = 36.48 N/m

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