A 1052 kg boulder slides down a sloping hillside during a landslide. The hillside slopes at an angle of 3 degrees to the horizontal, and the frictional forces on the boulder add up to 205 newton.What is the boulder's acceleration down the slope? Give your answer to one decimal place.

Question

BlackTom AI · Accepted Answer

The problem asks for the acceleration of a boulder sliding down a slope with given mass, slope angle, and total frictional force. 
First, identify the forces along the slope: gravity components along the slope provide driving force, while friction resists motion. 
The downslope component of gravity is m g sin(theta). With m = 1052 kg, g ≈ 9.8 m/s^2, and theta = 3 degrees, sin(3°) ≈ 0.052336, so m g sin(theta) ≈ 1052 × 9.8 × 0.052336 ≈ 1052 × 0.5128 ≈ 539.5 N (but we will focus on the net acceleration per unit mass below). 
The total frictional force opposing motion is F_f = 205 N. 
Net force along the slope is F_net = m g sin(theta) − F_f. 
To find acceleration, a = F_net / m = g sin(theta) − (F_f / m). 
Compute the terms: g sin(theta) ≈ 9.8 × 0.052336 ≈ 0.5128 m/s^2, and F_f / m = 205 / 1052 ≈ 0.1948 m/s^2. 
Subtracting yields a ≈ 0.5128 − 0.1948 ≈ 0.3180 m/s^2. 
Rounding to one decimal place gives a ≈ 0.3 m/s^2. 
Thus, the boulder accelerates downslope at about 0.3 m/s^2. 
Note on inputs: since no multiple-choice options were provided, this analysis used the standard physics model for motion on an incline with kinetic friction, and the result aligns with the given numerical answer format.

A 1052 kg boulder slides down a sloping hillside during a landslide. The hillside slopes at an angle of 3 degrees to the horizontal, and the frictional forces on the boulder add up to 205 newton.What is the boulder's acceleration down the slope? Give your answer to one decimal place.

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