A 2.0-kg block slides on a frictionless 25degree In dined plane A force of 4.6 N

Question

A 2.0-kg block slides on a frictionless 25degree In dined plane A force of 4.6 N acting parallel to the Incline and up the incline is applied to the bloc is the acceleration of the block? 1.8 m/s^2 up the incline 1.8 m/s^2 down the Incline 2.3 m/s^2 up the Incline 2.3 m/s^2 down the incline A highway curve has a radius of 0.14 km and is unbanked. A car weighing 12 kN goes around the curve at a speed of 24 m/s without slipping. What is the magnitude of the horizontal force (in kN) of the road on the car? 12 38 5 19 The horizontal surface on which the block of mass m = 3.4 kg slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. If k = 2.0 kN/m for the spring, then how far (in cm) does the block compress the spring before coming to rest? 30 15 10 25 Which of the following are possible units of momentum? Ns kg m/s kg m/s^2 I. and III. II. I. and II. III.

Explanation / Answer

6)

net force on block = 2*9.8*sin(25) - 4.6 = 3.68 N

acceleration = 3.68/2 = 1.84 m/s^2 down the incline

7)

horizontal force = mv^2/r = (12kN/9.8)*(24)^2/(140) = 5 kN

8)

(1/2)mv^2 = (1/2)kx^2

x = v*sqrt(m/k) = 6*sqrt(3.4/(2*10^3)) = 0.2473 m

x = 25 cm

9)

momentum = m*v

kg.m/s or N.s

option (c) is correct

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.