What is the mass of an object that has an acceleration of 2.63m/s2 when a unbalanced force of 112N is applied to it?
1 answer:
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Answer:
6.57 m/s
Explanation:
First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement
F=kx; F=180(.3) = 54 N
Next from Newton's second law find the acceleration of the mass.
Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²
Now use the kinematic equation for velocity (or speed)
v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.
v₀=0, since the mass is at rest before we release it
a=72 m/s² (from above)
x₀=0 as the start position already compressed
x₂=0.3m (this puts the spring back to it's natural length)
v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²
v₂= = 6.57 m/s
Answer:
B 14.5 m/s to the east
Explanation:
We can solve this problem by using the law of conservation of momentum.
In fact, if the system is isolated, the total momentum of the system must be conserved.
Here the total momentum before the stuntman reaches the skateboard is:
where
M = 72.0 kg is the mass of the stuntman
v = 15.0 m/s is his initial velocity (to the east)
The total momentum after the stuntmen reaches the skateboard is:
where
m = 2.50 kg is the mass of the skateboard
v' is the final velocity of the stuntman and the skateboard
Since momentum must be conserved, we have
And solvign for v',
And since the sign is the same as v, the direction is the same (to the east).