an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and
1 answer:
Answer:
The skater's speed after she stops pushing on the wall is 1.745 m/s.
Explanation:
Given that,
The average force exerted on the wall by an ice skater, F = 120 N
Time, t = 0.8 seconds
Mass of the skater, m = 55 kg
It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :
The change in momentum is equal to the impulse delivered. So,
So, the skater's speed after she stops pushing on the wall is 1.745 m/s.
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To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.
Since the forces are balanced the Spring force is equal to the force of the weight that is
Where,
k = Spring constant
x = Displacement
m = Mass
g = Gravitational Acceleration
Re-arrange to find the spring constant
Just before launch the compression is 40cm, then from Potential Elastic Energy definition
Therefore the energy stored in the spring is 63.72J
Answer:
6.18 m/s
Explanation:
Roller skate collision
The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;
x-axis component form (+x east);
+
+
=
+
Ф
60 ·8 + 0 = (60 + 80)Ф
480 = 140Ф................. (I)
y-axis component form (+y north);
+
+
=
+
Ф
0 + 80.9 = (60 + 80)Ф
720=
140Ф
140Vf=Ф......................................(2)
Substituting (2) into (1) to give the angle;
480 = 720tan Ф
Ф = arctan(0.67) =33.69°.......................(3)
Evaluating (1) with (3) gives the velocity magnitude
480 = 140Vfsin 33.69°
Vf=6.18 m/s
note 1:
This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.