","comment":{"@type":"Comment","text":" First, find the applied force and then find the frictional force acting on the lug by sketching out a free-body diagram. Use the same to arrive at an appropriate expression to calculate the net force. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$5.0 N$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$9.8 N$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$10.35 N$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$4.65 N$$ ","position":2,"answerExplanation":{"@type":"Comment","text":"$$\\\\ $$ Given that the mass of the lug of apples is m = 10kg, and is pushed at a rate of $$ a = 1 ms^{-2}$$ the force applied on the crate to produce this acceleration will be: $$ \\\\ F^{\\prime} = ma = 10 \\times 1 = 10 N\\\\ $$ Given that the coefficient of friction between the lug and the ground is $$ \\mu = 0.75 $$ the frictional force acting on the lug will be: $$ \\\\ F_{r} = \\mu N\\\\ \\Rightarrow F_{r} = \\mu\\, mg\\, \\cos30^{\\circ}\\\\ \\Rightarrow F_{r}= 0.75 \\times 10 \\times 9.8 \\times 0.866\\\\ \\Rightarrow F_{r}= 63.65 N\\\\ $$ From the diagram, we see that the net force acting on the lug can be given as: $$ \\\\ F = F_r - F^{\\prime} – mg\\,\\sin 30^{\\circ}\\\\ \\Rightarrow F = 63.65 – 10 – (10 \\times 9.8 \\times 0.5)\\\\ \\Rightarrow F = 53.65 – 49\\\\ \\therefore F = 4.65 N$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}