Please show all steps and explain how you got the answer!! and include the Free

Question

Please show all steps and explain how you got the answer!! and include the Free body diagram!! ty!

OA 0.21 kg block is placed against a spring which is compressed by 8 cm. After being released it slides into the 3.4 m diameter loop-the-loop of fricitonless track. At the point shown (one quarter of the way around the loop), the direction of the total acceleration of the block is 32 deg. below the horizontal. What is the spring constant of the spring? Hint: draw FBD Please show all steps and explain how you got the answer!! and include the Free body diagram!! ty!

Explanation / Answer

let v is the speed at the given point.

r = d/2 = 3.4/2 = 1.7 m

tan(theta) = ay/ax

tan(32)   = g/(v^2/r)

0.625   = 9.8/(v^2/1.7)

==> v = sqrt(9.8*1.7/0.625)

= 5.16 m/s

let u is the speed of the block when the block leaves the spring.

v^2 - u^2 = 2*a*s

v^2 - u^2 = -2*g*r)

u = sqrt(v^2 + 2*g*r)

= sqrt(5.16^2 + 2*9.8*1.7)

= 7.74 m/s

let k is the spring constant

Aplly energy conservation

0.5*k*x^2 = 0.5*m*u^2

k = m*u^2/x^2

= 0.21*7.74^2/0.08^2

= 1967 N/m <<<<<<<<<<<------------------Answer

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