A box of mass 2 kg has initial speed 10m/s. friction is 2n find height when speed is 5m/s and g=10n/kg
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Let m be the mass of the box and θ the angle of elevation of the ramp. We can write
ma = mg sinθ + F
Retardation a = g sin θ + F/m
= (10 N/kg) x sin 60° + 2N/2 kg
= 9.66 m/s2
Initial velocity at the foot of the ramp v0= 10 m/s
Final velocity v = 5 m/s
Let h be the height at which the body reaches its final velocity.
v2 − u2 = 2ah
h = (v2 − u2)/2a
= (5x5 − 10x10)/(−9.66)
= 7.76 m
ma = mg sinθ + F
Retardation a = g sin θ + F/m
= (10 N/kg) x sin 60° + 2N/2 kg
= 9.66 m/s2
Initial velocity at the foot of the ramp v0= 10 m/s
Final velocity v = 5 m/s
Let h be the height at which the body reaches its final velocity.
v2 − u2 = 2ah
h = (v2 − u2)/2a
= (5x5 − 10x10)/(−9.66)
= 7.76 m
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