Question

A 5 kg block is sliding on a horizontal surface while being pulled by a child using a rope attached to the center of the block. The rope exerts a constant force of 29.48 N at an angle of \theta=θ = 30 degrees above the horizontal on the block. Friction exists between the block and supporting surface (with \mu_s=\:μ s = 0.2 and \mu_k=\:μ k = 0.1 ). What is the horizontal acceleration of the block?

Answer 1

Answer 2

Calculation:

1. Resolve the Applied Force:
   The rope pulls with a force $F=29.48\text{N}$ at $30^{\circ }$.

• Horizontal component: $F_x=F\cos (30^{\circ })=29.48\times 0.866=25.52\text{N}$.

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• Vertical component: $F_y=F\sin (30^{\circ })=29.48\times 0.5=14.74\text{N}$.

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2. Calculate Normal Force ($N$):
   The block's weight $W=mg=5\text{kg}\times 9.8{\text{m/s}}^2=49\text{N}$.
   The vertical force $F_y$ lifts the block slightly, reducing the normal force:

   $$N=W-F_y=49\text{N}-14.74\text{N}=34.26\text{N}\mathrm{.}$$

4. Determine Frictional Force ($f$):
   Kinetic friction acts because the block is sliding:

   $$f={\mu }_{k}N=0.1\times 34.26\text{N}=3.426\text{N}\mathrm{.}$$

6. Compute Net Horizontal Force ($F_{\text{net}}$):

   $$F_{\text{net}}=F_x-f=25.52\text{N}-3.426\text{N}=22.094\text{N}\mathrm{.}$$

8. Find Acceleration ($a$):
   Using Newton's Second Law:

   $$a=\frac{F_{\text{net}}}{m}=\frac{22.094\text{N}}{5\text{kg}}=4.4188{\text{m/s}}^2\approx 4.0{\text{m/s}}^2\mathrm{.}$$

Answer:
The horizontal acceleration of the block is 4.0 m/s².