2.) A block (weight 40 N) is at rest at the bottom of a frictionless inclined pl

Question

2.) A block (weight 40 N) is at rest at the bottom of a frictionless inclined plane that makes an angle
= 36.87 degrees with the horizontal. A spring (k = 1600 N/m) is compressed a distance 0.2 m and is used
to shoot the block up the inclined plane. How far along the inclined plane does the block go (measured
parallel to the incline)?

A.) 0.80 m
B.) 1.00 m
C.) 1.33 m
D.) 10.0 m
E.) 13.3 m

W=mg; 40N = m(9.8 m/s^2) m=4.1 kg

(1/2)kx^2 = (1/2) mv^2
v= 3.95 m/s

v^2 = 2a (x) x=distance
v^2 = 2a (cos 36.87)x
x=.99 m = 1m

I got answer B. Is this correct?

Explanation / Answer

yes ur correct.

good work keep it up

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