Question #173365

A box weighing 225 N accelerates along a flat surface under the action of a force of intensity 14.6 N at an angle of 19 °. Determine the horizontal component of the box acceleration. When calculating, ignore any friction effects between the box and the surface of the substrate.


1
Expert's answer
2021-03-23T11:13:15-0400

Let's first calculate the mass of the box:


m=Wg=225 N9.8 ms2=22.96 kg.m=\dfrac{W}{g}=\dfrac{225\ N}{9.8\ \dfrac{m}{s^2}}=22.96\ kg.

Then, let's find the horizontal component of the force acting on the box:


Fh=Fcosθ=14.6 Ncos19=13.8 N.F_h=Fcos\theta=14.6\ N\cdot cos19^{\circ}=13.8\ N.

Finally, we can find the horizontal component of the box acceleration:


ah=Fhm=13.8 N22.96 kg=0.6 ms2.a_h=\dfrac{F_h}{m}=\dfrac{13.8\ N}{22.96\ kg}=0.6\ \dfrac{m}{s^2}.

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