An ice cube of mass 50.0 g can slide without friction up and down a 25.0 degree slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 0.100 m . The spring constant is 25.0 N/m . When the ice cube is released, how far will it travel up the slope before reversing direction

d = 0.604 m

Explanation:

m = 50 g =0.05 kg, θ = 25 ° , x= 0.100 m, K 25.0 N/m

At the bottom P.E. = 0 J and

K.E. = 1/2 K x² = 1/2  × 25 N/m × (0.100 m)² = 0.125 J

at maximum height after which it will reverse direction

P.E. = mgh        (P.E. = K.E.)

h = K.E. /mg = 0.125 J / (0.05 kg × 9.81 m/s²)

h= 0.255 m

Now h = d sin θ

d = 0.225 m / sin 25

d = 0.604 m

true is the correct answer

hope i

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