Answer:
Explanation:
Remark
In general, these 3rd class levers are very inefficient. Because the force distance is smaller than the load distance, you need to pull upward with more force that the weight of the load. So whatever the load is, the force is going to be much greater.
The distances are always measured to the pivot unless you are asked something specific otherwise.
Givens
F = ?
weight = 6N
Force Distance = F*d = 0.5 m
Weight Distance =W*d1 = 2 m
Formula
F*Fd = W*Wd
Solution
F*0.5 = 6 * 2 Divide by 0.5
F = 12/0.5
F = 24 N upwards
The answer is the last option
Explanation:
Answer:
P = 33.6 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = forces [N]
m = mass = 14 [kg]
a = acceleration = 6 [m/s²]
![F = 14*6\\F = 84 [N]](https://tex.z-dn.net/?f=F%20%3D%2014%2A6%5C%5CF%20%3D%2084%20%5BN%5D)
In the second part of this problem we must find the work done, where the work in physics is known as the product of force by distance, it is important to make it clear that force must be applied in the direction of movement.

where:
W = work [J]
F = force = 84 [N]
d = displaciment = 40 [m]
![W = 84*40\\W = 3360 [J]](https://tex.z-dn.net/?f=W%20%3D%2084%2A40%5C%5CW%20%3D%203360%20%5BJ%5D)
Finally, the power can be calculated by the relationship between the work performed in a given time interval.

where:
P = power [W]
W = work = 3360 [J]
t = time = 100 [s]
Now replacing:
![P=3360/100\\P=33.6[W]](https://tex.z-dn.net/?f=P%3D3360%2F100%5C%5CP%3D33.6%5BW%5D)
The power is given in watts
<span>Ans : Initial E = KE = ½mv² = ½ * 1.2kg * (2.2m/s)² = 2.9 J
max spring compression where both velocities are the same: conserve momentum:
1.2kg * 2.2m/s = (1.2 + 3.2)kg * v → v = 0.6 m/s
which means the combined KE = ½ * (1.2 + 3.2)kg * (0.6m/s)² = 0.79 J
The remaining energy went into the spring:
U = (2.9 - 0.79) J = 2.1 J = ½kx² = ½ * 554N/m * x²
x = 0.0076 m ↠(a)</span>