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.


Expert's answer

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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