A vertical spring with a spring constant value of 4500 N/m is compressed 60 cm. A 5.5 kg pumpkin is placed on the end of the spr
2 answers:
15.0 m
Explanation:
We can solve the problem by using the law of conservation of energy.
At the beginning, all the energy of the system is stored in the elastic potential energy of the compressed spring, which is given by
When the spring is released, the pumpkin of mass m=5.5 kg is pushed upwards. All the energy is then converted into gravitational potential energy of the pumpkin when it reaches its maximum height h:
Re-arranging the equation, we can find h:
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Answer: Normal force, N = 141.64 Newton
Explanation:
All the forces acting on the system and described in free body diagram are:
1) gravitational pull in downward direction
2) Normal force in upward direction
3) External force of 40 N acting at an angle of 37° with the horizontal can be resolved in two rectangular components:
i) F Cos 37° along the horizontal plane in forward direction and
ii) F Sin 37° along the vertical plane in downward direction
Applying the Newton's second law, net forces in the vertical plane are:
Net force, f = N - (mg + F Sin 37°)
As there is no acceleration in the vertical plane hence, net force f = 0.
So,
N - (mg + F Sin 37°) = 0
Adding (mg + F Sin 37°) both the sides in above equation, we get
N = mg + F Sin 37°
N = 12 9.8 + 40
0.601 because (Sin 37° = 0.601)
N = 117.6 + 24.04
N = 141.64 Newton
Answer:
Explanation:
Height reached by the object after push off is given as
now we have
now we know that this push last for total distance of 0.18 m
so during the push we will have
now in terms of g = 9.81 m/s/s we have