Answer:
26945.6 ft⋅lbf
Explanation:
Volume of Right Circular Cone = pi*(radius^2)*(height/3)
Pi*(4)*(5/3) = 20.94 ft^3
Density = Mass / Volume
Mass = Density*Volume
Mass = (40)*(20.94)
Mass = 837.6 lb
Work = Force*Height
Force = Mass*Acceleration
Acceleration will be gravitational acceleration
Work = (837.6)*(32.17)*(1)
Work = 26945.6 ft⋅lbf
ANSWER:
D) centripetal acceleration.
STEP-BY-STEP EXPLANATION:
When a body performs a uniform circular motion, the direction of the velocity vector changes at every instant. This variation is experienced by the linear vector, due to a force called centripetal, directed towards the center of the circumference that gives rise to the centripetal acceleration.
Therefore, the answer is centripetal acceleration.
The work done occurs only in the direction the block was moved - horizontally. Work is given by:
W = F(h) * d
Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.
Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):
F(h) = F(app)cos(23)
Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:
N = mg + F(v) = mg + F(app)sin(23)
Now we can get down to business and solve for F(app) - as mentioned above:
F(h) = f
F(h) = uN
F(h) = u * (mg + F(v))
F(app)cos(23) = 0.20 * (33 * 9.8 + F(app)sin(23))
F(app) = 76.8
Now that we have F(app), we can find the exact value of F(h):
F(h) = F(app)cos(23)
F(h) = 76.8cos(23)
F(h) = 70.7
And now that we have F(h), we can find W:
W = F(h) * d
W = 70.7 * 6.1
W = 431.3
Therefore, the work done by the worker's force is 431.3 J. This also represents the increase in thermal energy of the block-floor system.
