A 1100 kg car travelling at 27 m/s starts to decelerate and come to a complete s
ID: 1477705 • Letter: A
Question
A 1100 kg car travelling at 27 m/s starts to decelerate and come to a complete stop in 578.0 m. What is the average braking force acting on the car? The figure shows two forces acting on an object. They have magntitudes F1 = 6.3 N and d2 =2.1 N. What third force will cause the object to be in equilibrium? A child on a sled starts fromrest at the top of a 15.0 slope. If the trip to thebottom takes 15.2 s, how long is the slope? Assume that frictional force may be neglected. What is the mass of an object that experience a gravitational force of 164 N near Earth's surface?Explanation / Answer
17) use your constant acceleration equation.
Vf^2 - Vi^2 = 2ax
(27 m/s)^2 - 0 = 2a(578 m)
a = 0.6306 m/s^2
Now, use Newton's 2nd Law:
F = ma = (1100 kg)(0.6306 m/s^2)
F = 693.69 N
18) the resultant force will be if magnitude sqrt( F1^2 + F2^2 + 2*F1*F2*cost) t is the angle between both these force vectors,here its 90degrees.
so,the resultant will be sqrt(6.3^2 + 2.1^2) = sqrt(44.1) = 6.64, directed from center at 45degrees to F1 in the direction of F2.
so,the force required to balance this will be opposite to the resultant.
If suppose, F1 is along west, F2 along north, then resultant is in north-west, and force to cause this to be in equilibrium will be opposite to resultant in south-east)
19) From a free body diagram the force along the ramp is m*g*sin(15)
this equals m*a
so a = g*sin(15) = 9.80*sin(15) = 2.536m/s^2
Now apply kinematic eqn x = vx0*t + 1/2*a*t^2 where vx0 = 0
so x = 1/2*2.536m/s^2*(15.2s)^2 = 293m
20) You need to use the Newton's second law of motion such that:
F=ma =mg
Since the problem provides the value of the force and considering g=9.8 m/s2 yields:
m=164/9.8=16.73
Hence, evaluating the mass of the object under the given conditions yields m=16.73 kg .
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