We can start by finding the horizontal component of the force that the person exerts on the lawn mower using the given angle. We can call this horizontal force F_x.
F_x = F * cos(45)
F_x = 70 * cos(45)
F_x ≈ 49.5 N
We know that the retarding force acting on the lawn mower is 70*cos(45), which is equal to the horizontal component of the applied force (F_x) when the lawn mower is moving at a constant speed. So, the net force acting on the lawn mower when it's accelerating is the same as when it's moving at a constant speed, which is about 49.5 N.
Now we can use Newton's second law of motion, F_net = m * a, to find the acceleration (a) of the lawn mower.
49.5 = 16 * a
a ≈ 3.1 m/s^2
Now that we have the acceleration, we can use the following kinematic equation to find the final velocity (v) of the lawn mower after 2 seconds of acceleration, starting from rest (initial velocity, u = 0):
v = u + a*t
v = 0 + (3.1 * 2)
v ≈ 6.2 m/s
However, the problem states that the final velocity of the lawn mower should be 1.2 m/s. So, we need to adjust the net force to achieve this final velocity.
We can use the same kinematic equation with the desired final velocity, v = 1.2 m/s:
1.2 = 0 + a*2
a ≈ 0.6 m/s^2
Now we can use Newton's second law to find the required net force:
F_net = m * a
F_net = 16 * 0.6
F_net ≈ 9.6 N
Since the required net force is less than the initial net force (49.5 N), we need to find the difference in force that the person must exert to achieve the desired final velocity.
∆F = 49.5 - 9.6
∆F ≈ 39.9 N
Therefore, the person must exert an additional force of about 39.9 N on the lawn mower to accelerate it from rest to 1.2 m/s in 2 seconds.
A person pushes a 16.0 kg lawn mower at constant speed with a force of 70.0 N directed along the handle, which is at an angle of è = 45.0° to the horizontal. Calculate the force the person must exert on the lawn mower to accelerate it from rest to 1.2 m/s in 2.0 seconds (assuming the same retarding force which is 70cos45).
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