1)How much work is done by a person lifting a 2 kg object from the bottom of a well at a constant speed of 2m/s for 5 s. Is the answer mgh=2*9.8*10m
2)A 2kg block slides on a rough horizontal surface. A force (magnitude = 4N) acting parallel to the surface is applied to the block. The magnitude of the block¡¦s acceleration is 1.2m/s. If F is increased to 5N the magnitude of the block¡¦s acceleration is? I thought the answer would be a=F/m =5/2 but I think I am suppose to use the 1.2m/s and 4N in some way.
yes on 1)
2) YOu have to find friction first.
applied force - friction = m*a
so you know applied force 4N, m=2kg, and you are given a=1.2 so you can find friction.
Now friction force depends on mass, so if you increase F to 5N
use the same equation to find acceleration, youj know mass, and frictiona.
1- is incorrect lmfao
l m a o
1) To find the work done by a person lifting a 2 kg object, you can use the formula:
Work (W) = force (F) x displacement (d) x cos(theta)
In this case, the force exerted is equal to the weight of the object, which is given by the formula:
Force (F) = mass (m) x gravitational acceleration (g)
Substituting the given values:
F = 2 kg x 9.8 m/s^2
Now, let's calculate the work done:
W = F x d x cos(theta)
Since the object is lifted at a constant speed and the displacement is vertical, the angle between the force and displacement (theta) is 0 degrees. So, cos(theta) = 1.
Given: d = 2 m (constant speed x time)
W = (2 kg x 9.8 m/s^2) x (2 m) x (1)
W = 39.2 Joules
Therefore, the work done by the person lifting the 2 kg object is 39.2 Joules.
2) To calculate the magnitude of the block's acceleration when a force of 4 N is applied, you can use Newton's second law of motion:
F = m x a
Given: F = 4 N and m = 2 kg
Substituting these values into the formula:
4 N = 2 kg x a
Solving for a:
a = (4 N) / (2 kg)
a = 2 m/s^2
Therefore, when a force of 4 N is applied, the magnitude of the block's acceleration is 2 m/s^2.
Now, let's calculate the magnitude of the block's acceleration when the force is increased to 5 N.
Using the same formula:
5 N = 2 kg x a
Solving for a:
a = (5 N) / (2 kg)
a = 2.5 m/s^2
Therefore, when the force is increased to 5 N, the magnitude of the block's acceleration is 2.5 m/s^2.