This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **ferri (lf7249) – HW6 – mackie – (10611) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 3 . 6 kg block initially at rest is pulled to the right along a horizontal surface by a constant, horizontal force of 19 . 4 N. The coefficient of kinetic friction is 0 . 139. The acceleration of gravity is 9 . 8 m / s 2 . Find the speed of the block after it has moved 3 . 21 m. Correct answer: 5 . 08442 m / s. Explanation: We have to use the equation Δ K =- f s (1) , to calculate the change in the kinetic energy, Δ K . The net force exerted on the block is the sum of the applied 19 . 4 N force and the frictional force. Since the frictional force is in the direction opposite the displacement, it must be subtracted. The magnitude of the frictional force is f = μ N = μmg . Therefore the net force acting on the block is F net = F- μmg = 19 . 4 N- (0 . 139) (3 . 6 kg) (9 . 8 m / s 2 ) = 14 . 4961 N . Multiplying this constant force by the dis- placement, and using equation (1), we obtain Δ K = F net s = (14 . 4961 N) (3 . 21 m) = 46 . 5324 J = 1 2 mv 2 , since the initial velocity is zero. Therefore, v f = radicalbigg 2 Δ K m = radicalBigg 2 (46 . 5324 J) 3 . 6 kg = 5 . 08442 m / s . 002 10.0 points You leave your 50 W portable color TV on for 7 hours each day and you do not pay attention to the cost of electricity. If the dorm (or your parents) charged you for your electricity use and the cost was $0 . 1 / kW · h, what would be your monthly (30 day) bill? Correct answer: 1 . 05 dollars. Explanation: Let : P = 50 W and t = 7 h / day , The energy consumed in each day is W = P t = (50 W) (7 h / day) · kW 1000 W = 0 . 35 kW · h / day . In 30 days, you would use (30 day) (0 . 35 kW · h / day) = 10 . 5 kW · h , which would cost you (10 . 5 kW · h) ($0 . 1 / kW · h) = $1 . 05 . 003 10.0 points A car of weight 3100 N operating at a rate of 207 kW develops a maximum speed of 41 m / s on a level, horizontal road. Assuming that the resistive force (due to friction and air resistance) remains constant, what is the car’s maximum speed on an incline of 1 in 20; i.e. , if θ is the angle of the incline with the horizontal, sin θ = 1 / 20 ? Correct answer: 39 . 7788 m / s. Explanation: ferri (lf7249) – HW6 – mackie – (10611) 2 If f is the resisting force on a horizontal road, then the power P is P = f v horizontal , so that f = P v h = (2 . 07 × 10 5 W) (41 m / s) = 5048 . 78 N . On the incline, the resisting force is F = f + mg sin θ = f + W 20 = P v h + W 20 . And, F v = P , so v = P F = P P v h + W 20 = (2 . 07 × 10 5 W) (2 . 07 × 10 5 W) (41 m / s) + (3100 N) 20 = 39 . 7788 m / s ....

View
Full
Document