Question

1) A motorcycle's engine delivers a top power of 12*sqrt(v), the result being in units of horsepower when the velocity is in miles/hr. Since the motorcycle's velocity isn't changing, most of this power is going toward overcoming air resistance plus a force related to squashing the tires and road (see section 10-10 of the 4th edition of Giancoli or 10-12 of the 3rd edition if you are interested). A) How much work does the engine do in 100 milliseconds at 60 and 20 miles/hr at top power (assume constant speed during the 100ms)? B) What is the forward frictional force the road exerts on the tires at 60 and 20 miles/hr? C) What are the units of the constant "12" in the expression for the engine's top power?

Answer 1

When u=20 miles/hr, the power is: P= 12,50 P = 12/20 = 53.67 hp The work done by the engine is: W = Pxt w = 63.hp)15W) 7 x[acomshwa mrad) W = 400185 Ws W = 4001-85J When u= 60 miles/ho, the power is: P=12.60 = 92-95 hp The work done by the engine is: W =pxt W =[92-95 hp) x/145.7Wlx.com) 13 7 6- W=69313 At = 20 miles/hr, the force is! p=fyu (53-67 hp)x(745-7W) - ny)x(0.447 m 2 11hp 1 miles/h) TF = 4476.7N At u= 60 miles/ho, the force is : P=FxU (92.95 hp) x(745.4W) = -x (co miles/hx )x(0.447 m/s 97 F = 2584.37N P-t e unit of 12 = unit of P/unit of me = hp/Amiles/ho=hp&hi/smiles =hphics no Was 1972 -