Chapter 6 Work and Energy
What Is The Minimum Work Needed To Push A 950. In order to calculate the minimum work required. The work is equal to m g.
Web next, we use these distances to find the horizontal and vertical component of work. Web what is the minimum work needed to push a 950 kg car 810 m up along a 9.0° incline? The work is equal to m g. The component of the weight. W x = (850) (9.81) (4951.46) = 41287749.21 j w y = (850) (9.81) (740) = 6170409. A) ignore friction b) assume the effective coefficient of friction retarding the car is 0.25. Wc = mg = 950kg * 9.8n/kg 9310n. Givens (lowest number of digits: There is one force pushing the car down along the slope: Fp = 9310sin8 = 130n.
In order to calculate the minimum work required. Web what is the minimum work needed to push a 950 kg car 810 m up along a 9.0° incline? Web next, we use these distances to find the horizontal and vertical component of work. Signs feed at times the distance the mass. In this question, we will be solving for the break work needed to push the 950 kg car so we know that right is equal to the force of light times the these studs. Givens (lowest number of digits: We're told that this is a 950 kg car. W x = (850) (9.81) (4951.46) = 41287749.21 j w y = (850) (9.81) (740) = 6170409. Web minimum work, in this case, is the product of the weight of the car, w = 950 kg * 9.81 m'/s^2, by the height 710 m minimum work = 950*9.81*710 = 6616845 j (joules). There is one force pushing the car down along the slope: Web the minimum work needed is 1456.4 j for 1 meter of ramp explanation:
