A 0.52 kg object is at rest. A 3.45 N force to the right acts on the object during a time interval of 1.55 s.
A 0.52 kg object is at rest. A 3.45 N force to the right acts on the object during a time interval of 1.55 s. What is the velocity of the object at the end of this time interval? Answer in units of m/s.
At the end of this interval, a constant force of 4.41 N to the left is applied for 3.38 s. What is the velocity at the end of the 3.38 s? Answer in units of m/s.
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We assume that under the influence of the forces applied the object travels horizontally and there are no frictional forces to be overcome for this movement.
We know force is equal to mass multiplied by acceleration, that is:
f = m*a
or a = f/m
Where f = force, m = mass and a = acceleration
Let f1 be the initial force applied to the right and a1 the resultant acceleration. Then:
a1 = f1/m = 3.45/0.52
as the force is applied for 1,55 s (t1) the acceleration will continue for that much time.
Therefore final velocity v1 at the end of 1.55 s is given by formula:
v1 = a1*t1 = (3.45/0.52)*1.55 = 10.2836 m/s^2 to the right (approximately)
To find the velocity after the force o5 4.41 N is applied to the left for 3.38 s.
a2 = f2/m = -4.41/0.52
and velocity after after 3.38 seconds
v2 = v1 + a2*t2 = 10.2836 + (-4.41/0.52)*3.38 = 10.2836 + (28.665)
= -18.3814 m/s^2 to the right
or 18.3814 m/s^2 to the left
The acceleration on account of the force acting for 1.55s is equal to the force/mass of the object. or a= 3.45N/0.52.
Since the initial velocity u,zero(object was at rest), the velocity , v at the end of t seconds is given by:
v=u+at=0+(3.45/0.52)(1.55)=10.2837m/s to the right.
Taking now the initial velocity, u =10.2837m/s and a force of 4.41N in the opposite direction is applied. So the acceleration is force /mass =4.41N/0.52 and t=3.38s.
Therefore, v=u+at = 10.2837-(4.41/0.52)(3.38)= -18.3813 m/s, the -ve indicating the direction of the velocity is left at the end of 3.38 s
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