A 3000-kg truck accelerates from 10 m/s to 30 m/s in 8 s. The net force on the truck is

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BlackTom AI · Accepted Answer

The problem gives mass, initial and final speeds, and the time interval, so we can determine acceleration and then net force.
First, compute the acceleration from Δv = v_f − v_i = 30 − 10 = 20 m/s. Over a time interval Δt = 8 s, the acceleration a = Δv/Δt = 20/8 = 2.5 m/s^2.
Next, apply Newton's second law in its kinematic form: F_net = m a. With mass m = 3000 kg, F_net = 3000 × 2.5 = 7500 N.
Now evaluate the answer choices to see which aligns with this calculation.
Option A: 15,000 N. This would require an acceleration of F/m = 15000/3000 = 5 m/s^2, which is inconsistent with the given Δv and time, so this is too large.
Option B: 7500 N. This matches the computed net force using F = m a with a = Δv/Δt = 2.5 m/s^2, so this is consistent with the data.
Option C: 765 N. This is far too small for a 3000-kg mass undergoing the stated speed change; it would imply an acceleration of only 0.255 m/s^2, not supported by the given Δv/Δt.
Option D: 11,250 N. This would correspond to an acceleration of F/m = 11250/3000 = 3.75 m/s^2, which does not equal the acceleration derived from the speed change over the time interval.
In summary, only the calculation F = m a with a = Δv/Δt = 2.5 m/s^2 yields a net force of 7500 N, while the other options conflict with either the mass–acceleration relationship or the given kinematics.

Win25_PHYS010_10428 EC Quiz Ch.3-4

A 3000-kg truck accelerates from 10 m/s to 30 m/s in 8 s. The net force on the truck is

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