Question

A crate of mass m= 25.9 kg is in the bed of a pickup truck with the tailgate closed. there is negligible friction between the horizontal bed and the crate. the tailgate is vertical and contacts the back side of the crate. the truck accelerates forward on horizontal ground at a rate a= 3.8 m/s². what is the magnitude, in newtons, of the total force exerted on the crate by the truck?

A) 0 N
B) 25.9N
C) 250N
D) 631N

Answer 1

Using Newton's second law of motion, the total force exerted on the crate by the truck is calculated to be 98 N, which is not listed in the provided answer options.

Step-by-step explanation:

To determine the magnitude of the total force exerted on the crate by the truck, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = ma). Since the crate has a mass (m) of 25.9 kg and the truck accelerates at 3.8 m/s², the total force exerted on the crate by the truck can be calculated as follows:

F = ma = (25.9 kg)(3.8 m/s²) = 98.42 N

However, since the force has to be an integer value in Newtons, we round it to the nearest whole number, making it 98 N. So, the correct answer is that the total force exerted on the crate by the truck is 98 N, which is not one of the options provided. All the provided options are incorrect based on the given question parameters.