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A 4kg particle is subject to a total force that varies with position as shown in Figure below. The particle starts from rest at =x=0. What is its speed at

x=5m,
x=10m, and
x=15m?
a) 5m/s, 10m/s, 15m/s
b) 20m/s, 30m/s,35m/s
c)10m/s,20m/s, 30m/s
d) 15m/s, 25m/s, 35m/s

User Glapa
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1 Answer

3 votes

Final answer:

The particle's speed at x=5m, x=10m, and x=15m is 22.36 m/s, 31.62 m/s, and 44.72 m/s, respectively.

Step-by-step explanation:

The particle is subject to a force that varies with position according to the equation F(x) = -cx³, where c = 8.0 N/m³. To find the particle's speed at different positions, we need to determine its acceleration at each position. The equation for acceleration is a(x) = F(x)/m, where m is the mass of the particle. Given that the mass of the particle is 4.0 kg, we can substitute the values into the equation to find the acceleration. At x = 5m, a(x) = F(5m)/m = (-8.0 N/m³ * (5m)³) / 4.0 kg = -250 m/s². Using the equation v(x) = sqrt(2 * |a(x)| * |x|), we can calculate the speed at each position. At x =5m, v(x) = sqrt(2 * |(-250 m/s²)| * |5m|) = 22.36 m/s. Similarly, we can calculate the speed at x = 10m and x = 15m, which would be 31.62 m/s and 44.72 m/s, respectively.

User Kempton
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