phy A person pushes a 15 kg lawn mower at constant speed with a force of F = 90

Question

phy

A person pushes a 15 kg lawn mower at constant speed with a force of F = 90 N directed along the handle, which is at an angle of 45 to the horizontal (see Figure 1). Calculate the normal force exerted vertically upward on the mower by the ground Two objects are connected by a light string that passes over a frictionless pulley. The coefficient of kinetic friction between the 4-kg object and the surface is 0.30. Find the acceleration two objects. See Figure 2. A block is given an Initial speed of 5.0 m/s up the 20 plane shown In Figure 3. (a) How far up the plane will it go? (b) How much time elapses before it returns to its starting point? Ignore friction The asteroid lcarus-2, though only a few hundred meters across, orbits the Sun like the planets. Its period is 425 days. What is its mean distance from the Sun? A hypothetical planet has a mass 1.66 times that of Earth, but the same radius. What is g near its surface? How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius 85 m at a speed of 95 km/h

Explanation / Answer

1) sum forces in the y

N - mg - 90*sin(45) = 0

N = 15*9.81+90*sin(45 degrees)=210.8 N

2) Mg - friction = ( M + m) a

7*9.81 - 0.3*4*9.81 = 11*a

a=(7*9.81 - 0.3*4*9.81 )/11= 5.17

3)

a = g sin theta = 9.81*sin(20 degrees)=3.355

v^2 = v0^2 + 2 a x

x = 5^2/(2*3.355)=3.725 m

b) time up = time down

v = v0 + at

t = 5/3.355=1.49

so total time = 2.98 s

4) 4 pi^2/T^2 = G M/R^3

4*pi^2/(425*24*60*60)^2 = 6.67E-11*1.989E30/r^3

r=1.65E11 m

5)

a = G M/r^2 = G 1.66 Me/re^2 = 1.66 g = 1.66*9.81=16.28 m/s^2

6)

F = mv^2/r

u m g = mv^2/r
u = v^2/(gr)

95 km/hr = 26.39 m/s

u = 26.39^2/(9.81*85)=
0.835

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