Answer:
The work done on the canister by the 5.0 N force during this time is
54.06 Joules.
Explanation:
Let the initial kinetic energy of the canister be
KE₁ = =
= 19.44 J in the x direction
Let the the final kinetic energy of the canister be
KE₂ = =
= 73.5 J in the y direction
Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final
= 73.5 J - 19.44 J = 54.06 J
Answer:
Explanation:
Applied force, F = 18 N
Coefficient of static friction, μs = 0.4
Coefficient of kinetic friction, μs = 0.3
θ = 27°
Let N be the normal reaction of the wall acting on the block and m be the mass of block.
Resolve the components of force F.
As the block is in the horizontal equilibrium, so
F Cos 27° = N
N = 18 Cos 27° = 16.04 N
As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .
The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)
The vertically downward force acting on the block is mg - F Sin 27°
= mg - 18 Sin 27° = mg - 8.172 ... (2)
Now by equating the forces from equation (1) and (2), we get
mg - 8.172 = 6.42
mg = 14.592
m x 9.8 = 14.592
m = 1.49 kg
Thus, the mass of block is 1.5 kg.
