Answer:
Work = 0.36N
Explanation:
Given
Force = 12N
Distance = 0.03m
Weight = 0.05kg
Required
Determine the work done
Workdone is calculated as thus;
Work = Force * Distance
Substitute 12N for Force and 0.03m for Distance
Work = 12N * 0.03m
Work = 0.36Nm
Using proper S.I units
Work = 0.36N
Hence, work done by the spring on the ball is 0.36N
Answer:
The graph appears to be in error.
The actual figure appears to be a rhombus with sides of 5 and 15 with a height of 5
The work done (F * S) is the area of the rhombus
1/2 * (5 +15) * 5 = 50 J
Answer:
a) 1.76 m/s²
b) 75.5°
Explanation:
1. linear(tangential) acceleration a:
a = 0.44m/s²
2. speed v:
v(t) = at
3. radial acceleration α on a track with radius r:
α(t) = v²/r = (at)²/r
α(18)= (0.44*18)²/37 = 1.7 m/s²
4. total acceleration |a|:
|a| = √(a²+α²)=1.76m/s²
5. angle Ф with v:
tanФ =(α/a) => Ф = 75.5°
Answer:
b. 0.20 m/s.
Explanation:
Given;
initial mass, m = 0.2 kg
maximum speed, v = 0.3 m/s
The total energy of the spring at the given maximum speed is calculated as;
K.E = ¹/₂mv²
K.E = 0.5 x 0.2 x 0.3²
K.E = 0.009 J
If the mass is changed to 0.4 kg
¹/₂mv² = K.E
mv² = 2K.E
Therefore, the maximum speed is 0.20 m/s