Answer:
F = 200 N
Explanation:
Given that,
The mass suspended from the rope, m = 20 kg
We need to find the resultant force acting on the rope. The resultant force on the rope is equal to its weight such that,
F = mg
Where
g is acceleration due to gravity
Put all the values,
F = 20 kg × 10 m/s²
F = 200 N
So, the resultant force on the mass is 200 N.
Answer:
Newton's first law: An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force. ... An object sliding across a table or floor slows down due to the net force of friction acting on the object.
Explanation:
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Answer:
a) 19440 km/h²
b) 10 sec
Explanation:
v₀ = initial velocity of the car = 45 km/h
v = final velocity achieved by the car = 99 km/h
d = distance traveled by the car while accelerating = 0.2 km
a = acceleration of the car
Using the kinematics equation
v² = v₀² + 2 a d
99² = 45² + 2 a (0.2)
a = 19440 km/h²
b)
t = time required to reach the final velocity
Using the kinematics equation
v = v₀ + a t
99 = 45 + (19440) t
t = 0.00278 h
t = 0.00278 x 3600 sec
t = 10 sec
Answer:
A) 89.39 J
B) 30.39J
C) 23.8 J
Explanation:
We are given;
F = 30.2N
m = 3.5 kg
μ_k = 0.646
d = 2.96m
ΔEth (Block) = 35.2J
A) Work done by the applied force on the block-floor system is given as;
W = F•d
Thus, W = 30.2 x 2.96 = 89.39 J
B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;
ΔEth = μ_k•mgd
Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J
Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.
Thus,
ΔEth = ΔEth (Block) + ΔEth (floor)
Thus,
ΔEth (floor) = ΔEth - ΔEth (Block)
ΔEth (floor) = 65.59J - 35.2J = 30.39J
C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;
W = K + ΔEth
Therefore;
K = W - ΔEth
K = 89.39 - 65.59 = 23.8J
