A block rests on a horizontal frictionless surface. A string is attached to the block, and is pulled with a force of 60.0 N at an angle θ above the horizontal. After the block is pulled through a distance of 2.00 m, its speed is 2.70 m/s, and 43.0 J of work has been done on it.
A)What is the angle θ?
B)Calculate the mass of the block.
A. Work = F*cosA * d = 43 J.
60*cosA * 2.0 = 43
120*cosA = 43
cosA = 0.35833
A = 69o
B. a = (V^2-Vo^2)/2d
a = (2.7^2-0)/4 = 1.82 m/s^2.
F*cos69 = m*a
60*cos69 = m*1.82
m = 60*cos69/1.82 = 11.80 kg.
To find the angle θ, we can use the work-energy theorem. The work done on an object is equal to the change in its kinetic energy.
The work done on the block is given as 43.0 J. The change in kinetic energy can be calculated using the formula:
ΔKE = KE_final - KE_initial
Given that the final speed is 2.70 m/s, and assuming the initial speed was 0 (as the block starts from rest), we can calculate the change in kinetic energy as:
ΔKE = 0.5 * m * (vf)^2 - 0.5 * m * (vi)^2
Where m is the mass of the block, vf is the final velocity (2.70 m/s), and vi is the initial velocity (0 m/s).
Since ΔKE is equal to the work done (43.0 J), we can equate the two expressions and solve for the angle θ.
A) θ = arccos((2 * W) / (m * (vf)^2))
Where W is the work done and vf is the final velocity.
To calculate the mass of the block, we need to rearrange the equation above. Let's solve for m:
m = (2 * W) / ((vf)^2 * cos(θ))
B) m = (2 * 43.0 J) / ((2.70 m/s)^2 * cos(θ))
Please note that we still need to find the value of θ to calculate the mass accurately.