Using Hooke's law, F subscript s p r i n g end subscript equals k increment x, find the elastic constant of a spring that stretc
2 answers:
Let's rewrite Hook's law:
where
F is the force
k is the spring's constant
x is the compression/elongation of the spring
Neglecting the sign (since we are not interested in the direction of the force), we can re-arrange the formula and use the data of the problem to find the spring constant:
Therefore, the correct answer is B) 2 N/cm.
where
F is the force
k is the spring's constant
x is the compression/elongation of the spring
Neglecting the sign (since we are not interested in the direction of the force), we can re-arrange the formula and use the data of the problem to find the spring constant:
Therefore, the correct answer is B) 2 N/cm.
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Answer:
The velocity of block = 0.188
Explanation:
Mass m = 5.6 kg
k = 1040
= 0.26
0.035 m ,
= 0
0.02 m
From work energy theorem
--------- (1)
Kinetic energy
------- (1)
Potential energy
------- (2)
Work done
W = F.s ------ (3)
From Newton's second law
= mg
= 5.6 × 9.81 = 54.9 N
Friction force = 0.4 × 54.9 = 21.9 N
Now the work done by the friction
= - 21.9 × 0.015
= - 0.329 J
Now kinetic energy
At point 1
0.637 J
At point 2
Potential energy
J
From equation (1) we get
0 + 0.637 - 0.329 = 2.8 + 0.208
2.8 = 0.1
0.188
This is the velocity of block.
Answer:
Explanation:
Let the total force be F
The force that produces motion in the x direction is given as,
Let the frictional force be given as:
Thus, the net force is given as: