A 5.3-kg block is pushed 2.3 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of θ = 30° with the horizontal, as shown in the figure below. If the coefficient of kinetic friction between block and wall is 0.30, determine the following.
(a) the work done by F
J
(b) the work done by the force of gravity
J
(c) the work done by the normal force between block and wall
J
(d) By how much does the gravitational potential energy increase during the block's motion?
J
im getting wrong answer for 1) i got 216.2 J but that is incorrect
2) im also getting wrong when i plug in everything.
3) is 0
4) is 119.50
so can u please look at 1 and 2 and help me
To determine the work done by each force and the increase in gravitational potential energy, we need to understand the concept of work and apply it to the given scenario.
Work (W) is defined as the product of force exerted on an object and the displacement of the object in the direction of the force. Mathematically, it can be calculated using the formula:
W = F * d * cos(θ)
where W is the work done, F is the magnitude of the force, d is the displacement, and θ is the angle between the force and displacement vectors.
Let's calculate each part step by step:
(a) Work done by force F:
Given force F = F_applied at an angle of θ = 30° with the horizontal.
We need to find the work done by this force.
W_F = F * d * cos(θ)
= F * d * cos(30°) [Substituting given values]
= F * d * (√3/2)
= F * 2.3 * (√3/2) [Substituting the given displacement, d = 2.3m]
= F * 2.3 * 0.866 [Approximating the value of (√3/2) as 0.866]
Therefore, the work done by force F is F * 2.3 * 0.866 J.
(b) Work done by the force of gravity:
The force of gravity acts vertically downward on the block.
To calculate the work done by this force, we need to find the vertical component of the displacement, as the gravitational force and vertical displacement are in the same direction.
W_gravity = m * g * d_vertical [Using the formula W = F * d * cos(θ)]
= m * g * d * sin(θ) [Using the trigonometric relationship]
Given m = 5.3 kg, g = 9.8 m/s² (acceleration due to gravity), and θ = 30°.
We need to find the vertical displacement, d_vertical.
Since the block is moved up a vertical wall, the vertical displacement would be equal to the total displacement, d = 2.3m.
W_gravity = 5.3 kg * 9.8 m/s² * 2.3 m * sin(30°)
= 5.3 * 9.8 * 2.3 * 0.5
= 60.114 J [Approximating the value]
Therefore, the work done by the force of gravity is approximately 60.114 J.
(c) Work done by the normal force:
The normal force is responsible for exerting a force perpendicular to the displacement, resulting in no work done. Therefore, the work done by the normal force is zero (0 J).
(d) Increase in gravitational potential energy:
The increase in gravitational potential energy is equal to the negative work done by the force of gravity. Mathematically, we can calculate it as:
ΔPE_gravity = -W_gravity
Given the value of W_gravity as 60.114 J:
ΔPE_gravity = - 60.114 J
Therefore, the increase in gravitational potential energy during the block's motion is approximately -60.114 J.
I Response Any Thing what You Do
a) Find F:
F = mg/(sin30 - mu cos30)
Work = F sin30 (2.3)
b) mg (2.3)
c) normal force can never do work
d) mgh (same as b)