A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.85 m/s2
1 answer:
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Answer:
Explanation:
Given that
Mass , m = 25 kg
We know that when body is in rest condition then static friction force act on the body and when body is in motion the kinetic friction force act on the body .That is why these two forces are given as follows
Static friction force ,fs= 165 N
Kinetic friction force ,fk = 127 N
If the body is moving with constant velocity ,it means that acceleration of that body is zero and all the forces are balanced.
Lets take coefficient of kinetic friction = μk
The kinetic friction is given as follows
fk = μk m g
Now by putting the values
127 = μk x 25 x 9.81
Therefore the value of coefficient of kinetic friction will be 0.51
Answer:
(a) the mechanical energy of the system, U = 0.1078 J
(b) the maximum speed of the object, Vmax = 0.657 m/s
(c) the maximum acceleration of the object, a_max = 15.4 m/s²
Explanation:
Given;
Amplitude of the spring, A = 2.8 cm = 0.028 m
Spring constant, K = 275 N/m
Mass of object, m = 0.5 kg
(a) the mechanical energy of the system
This is the potential energy of the system, U = ¹/₂KA²
U = ¹/₂ (275)(0.028)²
U = 0.1078 J
(b) the maximum speed of the object
(c) the maximum acceleration of the object