Answer: (a) The work done by the applied force on the block is 199.335J
(b) The increase in thermal energy of the floor is 48.013J
(c) The increase in the kinetic energy of the block is 81.322J
Explanation: Here is the complete question:
A horizontal force of magnitude 41.1 N pushes a block of mass 4.18 kg across a floor where the coefficient of kinetic friction is 0.594. (a) How much work is done by that applied force on the block – floor system when the block slides through a displacement of 4.85 m across the floor? (b) During that displacement, the thermal energy of the block increases by 70.0 J. What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block?
Explanation: Please see the attachments below
Answer:
Explanation:
Given:
Force, f = 41.1 N
Mass = 4.18 kg
Coefficient of kinetic friction, Uk = 0.594
Displacement, x = 4.85 m
Workdone, E = force × distance, x
= 41.1 × 4.85
= 199.335 J
= 0.2 kJ.
